2c:null 2d:["$","title",null,{"children":"Untitled | Daniel LaLiberte | Observable"}] 32:T54d,### Nested Intervals Another way to look Cantor's Diagonalization argument is by drawing a parallel with his Nested Closed Intervals argument. For each subsequent number in the enumeration, another digit will be added to the diagonal number, and the complement of that digit to the diagonal complement number. For each prefix of the diagonal complement number, we can define an interval around it bounded by the number followed by all 0s on the low side and the same number followed by all 1s on the high side. Each subsequent interval constructed from successively longer prefixes of the diagonal complement number will be nested inside the previous interval. The limit of this sequence of nested closed intervals based on prefixes of the diagonal complement digits is the complete diagonal complement number. The nested intervals argument is that the limit of the sequence of nested intervals, the diagonal complement number in this case, should be in the enumeration, but it is excluded by always choosing the subsequent intervals that exclude a number that is inside the preceding intervals So the diagonalization argument is directly analogous to the nested intervals argument, when framed in this way. This means that if we can refute the nested intervals argument, we effectively refute the diagonalization argument as well, and vice versa. 33:T5d5,// Generate one random number as first number and diagonal. // Generate prefixes of the diagonal complement up to the diagonal. function* generateOneRandomNumberAsDiagonal( numNumbers = 32, pregenerateXDepth, animate ) { const numbers = []; let index = 0; // Generate root of tree. const rootNode = new BinaryTreeNode(null, "0"); const randomDigits = generateNRandomDigits(numNumbers).slice(); // db('nRandomDigits', nRandomDigits); let nextDigit = randomDigits.shift(); // Generate the number for X let x = Real.construct(rootNode, index++, "X", [nextDigit]); let xComplementDigits = []; numbers.push(x); let parentNode = rootNode; while (numbers.length < numNumbers) { nextDigit = randomDigits.shift(); // Add nextDigit to curentNum x.addDigit(nextDigit); if (xComplementDigits.length > 0) { const newDigits = [...xComplementDigits, nextDigit]; const newNum = Real.construct(rootNode, index++, null, newDigits); // console.info('newNode', newNode); numbers.push(newNum); if (nextDigit == "0") { newNum.startOfInterval = true; } else { newNum.endOfInterval = true; } } // Append complement of next digit. xComplementDigits.push(nextDigit == "0" ? "1" : "0"); if (animate) { yield { tree: rootNode.traverseTree(), numbers }; } } // Return list of nodes by traversing the binary tree. return yield { tree: rootNode.traverseTree(), numbers }; }34:T48b,### Generate Nested Intervals around Diagonal Complement, Breadth-first This generator is a variation of the complete breadth-first generator that enumerates all real numbers between 0 and 1 that adds two enhancements which integrate features of the previous example that shows "Nested Intervals around the Diagonal Complement". 1. Use the first number as the diagonal. 2. Show nested intervals around the diagonal complement number. The number labelled 'X' is the first number and the diagonal number. The numbers labelled like '-X#' are approximations of the diagonal complement number. There are a few bugs to fix. The unlabelled interval boundaries should often not be interval boundaries since they appear to be out of order. And sometimes the innermost interval boundaries are wrong or missing. These bugs are probably due to the fact that each number, given its finite number of prefix digits, is only an approximation of the full number, so in a sense, each number represents a range of all possible extensions, from all 0s to all 1s. The problem would be simpler if we constrain each diagonal complement approximation to only last one step. 35:Tbc6,// Generate breadth-first numbers with first number along diagonal // And show intervals around diagonal complement number. // This could be refactored to be more clear. function generateIntervalsBreadthFirst(numNumbers = 64) { const numbers = []; let index = 0; // Generate root of tree. const rootNode = new BinaryTreeNode(null, "0"); const randomDigits = generateNRandomDigits(numNumbers).slice(); // db("randomDigits", randomDigits); let nextDigit = randomDigits.shift(); // Generate the number for X const x = Real.construct(rootNode.addDigit(nextDigit), index++, "X"); numbers.push(x); // Create initial xComplement with the other digit. const otherDigit = nextDigit == "0" ? "1" : "0"; let xComplement = Real.construct( rootNode.addDigit(otherDigit), index++, "-X0" ); numbers.push(xComplement); // Initial values of interval bounds. let lowerBoundValue = 0; let upperBoundValue = 1; if (nextDigit == "0") { lowerBoundValue = x.getValue(); } else { upperBoundValue = x.getValue(); } let diagonalComplementValue = xComplement.getValue(); const checkBounds = (num, digit) => { const value = num.getValue(); if (lowerBoundValue <= value && value <= upperBoundValue) { //if (value == diagonalComplementValue) { //} else if (value <= diagonalComplementValue) { lowerBoundValue = value; num.startOfInterval = true; num.endOfInterval = false; } else if (value > diagonalComplementValue) { upperBoundValue = value; num.startOfInterval = false; num.endOfInterval = true; } } }; checkBounds(x); checkBounds(xComplement); let depth = 1; while (numbers.length < numNumbers) { const currentNumbersLength = numbers.length; // Get nextDiagonalDigit const nextDiagonalDigit = randomDigits.shift(); for (let i = 0; i < currentNumbersLength; i++) { const currentNum = numbers[i]; const parentNode = currentNum.node; let nextDigit; if (i == 0 || i == depth) { // Add diagonal digit to first number and diagonal number nextDigit = nextDiagonalDigit; currentNum.addDigit(nextDigit); } else { // Add one digit to every current number after the first. nextDigit = currentNum.addNextDigit(); } checkBounds(currentNum, nextDigit); // Create a new number with the other digit. const otherDigit = nextDigit == "0" ? "1" : "0"; const newNode = parentNode.addDigit(otherDigit); const newNum = Real.construct(newNode, index++); numbers.push(newNum); if (currentNum == xComplement && nextDigit == nextDiagonalDigit) { xComplement = newNum; xComplement.setName("-X" + depth); diagonalComplementValue = newNum.getValue(); } checkBounds(newNum, otherDigit); } depth++; } // Return list of nodes by traversing the binary tree. return { tree: rootNode.traverseTree(), numbers }; }36:Teee,### Nested Closed Intervals Equivalent to Diagonalization From the above two examples, we can see that the breadth-first enumeration of the Reals implicitly creates a sequence of nested closed intervals around the diagonal complement number which is always excluded from the enumeration since that number is always different from the diagonal number at the diagonal digit. We make these nested closed intervals explicit in the visualization. When presented this way, Cantor's Nested Closed Intervals argument is equivalent to his Diagonalization argument, at least regarding this diagonal complement number. Whereas the Cantor's Diagonalization argument only applies to one number that is the complement of the diagonal number, his Nested Closed Intervals argument is generalized to apply to any number within a continuous range. As we saw previously in "Extension of Cantor's Diagonalization Argument to Include all Permutations of the Enumeration", the Diagonalization argument also applies to any number within a continuous range, if we allow permutations of the enumeration since every permutation of the same enumeration, which has its own diagonal complement number, will exclude every diagonal complement number from any permutation of the same enumeration. And in particular, this argument applies to any number already in the enumeration. But if we exclude each number in turn from the enumeration, we also notice that all permutations of the remaining numbers can still be used to exclude all the rest, and this results in a contradiction since we obviously started with an infinite number of numbers in the enumeration. This same line of reasoning applies to the Nested Closed Intervals argument. One number that is more clearly excluded from the enumeration is the diagonal complement number, and we can see how it is excluded by constructing the nested closed intervals around the approximations of that number. But, as Cantor argued, the nested closed intervals argument also applies to any number in the continuous range. We can see either directly, by choosing intervals that exclude any number, regardless whether it is in the enumeration, or by permuting the enumeration such that the diagonal complement number is excluded. Cantor's Nested Closed Intervals argument states that _any_ number within a continuous range can be excluded by constructing a sequence of closed intervals that excludes every number from the enumeration, and yet the sequence has a limit that is a single real number that ought to have been in the enumeration. But this same construction can be used to exclude any number that is otherwise in the enumeration. All we need to do is not use it in the construction of the sequence of nested closed intervals, and then we will discover that it is missing. Well, of course it is missing, you might say, because we chose not to include it. But the point is that we can't tell the difference between a number that we know is missing (because we explicitly removed it) from any other number that may or may not be missing. But, you might respond, any number that is not missing should eventually be used in the construction of the nested closed intervals. Well, this is not true if it happens to be the same as the diagonal complement number. (not finished...) including not only every number that is supposedly missing from the enumeration, but also any number that is in the enumeration. This makes the argument inherently contradictory. The refutation of his Diagonalization argument, presented earlier, makes it clear that the exclusion of any number from the enumeration, whether it is the diagonal complement or any other number that is already in the enumeration, is illusory. And so this same refutation applies to his Nested Closed Intervals argument.37:T682,### Is the Diagonal Complement Number in the Limit Set? One possible answer to this question of what happened to the diagonal complement number is that it only shows up at the limit. Since every finite-length prefix of the diagonal complement number can be used further down the enumeration, if we take the limit of the sequence of finite-length prefixes, we will get an infinite-length diagonal complement number. By transfinite induction, this infinite-length diagonal complement number _is_ in the sequence, but only at the index of infinity. But that is not a satisfactory answer because the diagonal complement number is forced to be beyond any finite position and thus _outside_ of the enumeration. The diagonal complement number _is_ in the _limit set_ of the enumeration, but the members of the limit set of a sequence are not necessarily the same as any of the members of the sequence. As another more familiar example, if we start with the sequence of fractions, 1/2, 1/4, 1/8, etc, every number in the sequence is greater than 0, so 0 is never in the sequence even though the limit of this sequence is 0. If we want to argue that the definition of the diagonal complement number is valid and it is actually included in the enumeration, we are essentially arguing that every infinite sequence of numbers should implicitly include every number in its limit set. That would mean, for example, that we only need to enumerate the rationals between 0 and 1 to effectively enumerate all reals that are in the limit set of that enumeration. Maybe we will get there, but I don't think we should make that leap of logic just yet, so let's look elsewhere.38:T49f,### Nested Intervals of Reals Must Include Infinitesimals at the Limit The limit of this sequence of nested intervals should contain a single number, at least according to the usual understanding of the "limit". But since the two bounds of the innermost interval will always be nearest neighbors, that means they must be _different_ numbers from each other and from the limit, in which case we either haven't really iterated to the limit yet, or it must mean that we need to interpret this concept of a limit differently and acknowledge that the bounds are only infinitesimally different from the limit. The usual meaning of the limit of the infinite sequence of closed intervals is that the bounds of the intersection of all these intervals are indistinguishable from the limit. But with our enumeration of reals, the nearest neighbors of the limit remain distinct from the limit and become infinitesimally close to the limit. So it appears we need to augment the real numbers with the hyperreal numbers in order to be complete. We will set aside this question for now, but another aspect of the flaw in the Nested Intervals argument remains, as explained in more detail below.39:T650,### Practically Indistinguishable, but Infinitesimally Distinct Bounds This process of constructing nested intervals while enumerating the real numbers can continue indefinitely, and note that at every step of the enumeration, the two bounds of the innermost interval will remain distinct. By transfinite induction, at the limit of the enumeration, the two bounds of the innermost interval will still be infinitesimally different from each other. While there is a real number that is the limit of this sequence of nested intervals, in the usual sense of a limit, how do we decide which of the two bounds we would use as the one limit? The two bounds are only infinitesimally different from each other, so if we can disregard infinitesimal differences, they are indistinguishable from each other. So why should one of the bounds be determined to be the real number, and the other is therefore that number plus or minus an infinitesimal? Wouldn't the reverse also be just as true? The fact that the two bounds of the innermost interval remain infinitesimally distinct from each other is one problem with Cantor's Nested Interval argument, but it is a problem that he inherited from the definition of limits, and the lack of inclusion of infinitesimals in the real numbers. The concept of limits allow us to leap over this final infinitesimal gap between the bounds of the innermost interval so we have grown comfortable with pretending that the gap never existed, or that it doesn't matter even if it does exist. But there is another significant problem with the Nested Intervals argument that we discuss next.3a:T87c,### Self-Referential Definition of Nested Intervals The essense of Cantor's Nested Intervals argument is that the limit of the sequence of nested closed intervals cannot be in the enumeration. There are three possibilities for what is happening with this sequence of nested closed intervals, two of which we can dispense with quickly. First, if the sequence of intervals is finite, then the final interval will still contain an infinite number of numbers that must have been missed, so this case would not occur unless the enumeration were incomplete. Second, if the sequence of nested intervals is infinite, but the bounds of the innermost interval at the limit of the sequence are still different, and if the difference between the bounds is a _finite_ number, then there must be an infinite number of finite numbers that have been missed between the two bounds, so this case could also not occur unless the enumeration were incomplete. However, if the bounds of the innermost interval are only _infinitesimally_ distinct, this is actually the situation described in the previous section. In this case, there are also an infinite number of numbers that can be squeezed between any two numbers even though they are only infinitesimally distinct, but all of these missing numbers are infinitesimals of a higher order. We believe this is the correct interpretation of what is happening with the nested closed intervals. The third possibility regarding the limit of the sequence of nested closed intervals is that the two bounds of the innermost interval are the same. In this case, the innermost interval must contain only this one number, the limit of the sequence. Cantor argues that this limit number could not be in the enumeration because otherwise the construction of the sequence of intervals would have encountered it earlier in the enumeration and used it for one of the bounds of the next interval. This is the contradiction that suggests the enumeration must be incomplete. But next we argue that the problem is not with the enumeration, but with the definition of the nested closed intervals combined with the notion of the limit of the sequence.3b:T93c,### The Limit of Nested Closed Intervals First, note that this third possibility would actually *never* occur in our enumeration process since the two bounds of the innermost interval will always be different though only infinitesimally distinct, as argued previously. That is, the supposition that the innermost interval of this sequence might contain _only_ the limit of the sequence must be wrong. Even if we assume that this third alternative is possible, we should note that the definition of the nested closed intervals is essentially self-referential. That is, each number we find inside the innermost known interval becomes one of the bounds of a subsequent nested interval in the sequence, so the sequence of intervals (as a whole) is defined self-referentially, in terms of the numbers contained in the intervals themselves. This self-referential definition of the sequence of intervals might have been OK, but it becomes a contradictory definition when combined with the notion of limits. Given the usual meaning of the limit of a sequence, at the limit of this sequence of nested closed intervals, there ought to be only one limit number, since we know that _every_ sequence of nested closed intervals _does_ have a limit, so this limit number exists. Furthermore, this limit number ought to have been included in any complete enumeration. Since any enumerated number we find inside the innermost interval, including this limit number, if it was in the enumeration, must be used as a bound of a subsequent interval, then if there is a limit number found, it must have not have been in the enumeration after all. But the real problem is not that the number must have been missed in the enumeration, but rather, that the sequence of nested closed intervals used here is defined in terms of the intervals themselves. The intervals are defined such that they do _not_ contain any previously visited numbers. By this definition, the limit number could not be included in the enumeration. The main problem with Cantor's argument is that we are led to believe that it should be possible for the limit of a sequence of nested closed intervals to be in the enumeration, whereas we effectively preclude, by the definition of the sequence of nested intervals, every possible number that we do enumerate from being inside the innermost nested interval.3c:T645,### Nested Intervals Argument Applies to Rationals Another severe problem with Cantor's nested intervals argument is that it applies to an enumeration of rational numbers as well, as long as we restrict the bounds of the nested intervals to rational numbers. In that case, then we should expect that the limit of the sequence of closed nested intervals with rational bounds would be a rational number as well, and so for the same reason that we might believe the argument regarding an enumeration of real numbers, we would have to conclude that the rationals must also be uncountable. Let's examine this refutation argument in more detail. We know that between any two distinct rational numbers, there are an infinite number of rational numbers, as well as an infinite number of irrational numbers. This is just as true of any two distinct irrational numbers. So when we are enumerating elements of the set of real numbers, including both rational and irrational numbers, we will always find an infinite number of rational and irrational numbers between any two real numbers. And this will also be true if we are enumerating only irrational numbers or only rational numbers. Except that the sequence of nested intervals with rational bounds might converge on an irrational number, but can a sequence of nested intervals with irrational bounds converge on a rational number? Only if, at the limit, the irrational factor becomes infinitesimally small. E.g. if Q is our rational limit, then the sequence with bounds being Q - pi / n and Q + pi / n would all be irrational, but they converge on Q.3d:T6ea,#### A Countable Infinity of Enumerations of Sets of Infinitesimals In the same way we subdivided the set of Reals between 0 and 1 into two subsets with half the range of numbers but the same cardinality, we can create N subsets with ranges of numbers that have boundary values of 0, 1/N, 2/N, ..., N-1/N, and 1. Every enumeration of any one of these subsets will have a diagonal complement number that should be a member of another subset, if we use binary notation and constrain N to be a power of 2. E.g. if N is 4, we will create four subsets; all numbers in the first subset will have leading "00" digits, and the complement numbers are all in the fourth subset with leading "11" digits. As N grows larger and larger, the number of subsets increases and the range of each subset decreases, while the cardinality of each subset remains the same (as all Reals). However as N approaches infinity, the number of subsets grows to infinity, and the range of each subset shrinks to 0, but the cardinality of each subset suddenly drops all the way from infinity down to 0. But wait, does that make sense? If we consider the possibility that an infinitesimal (a value that is smaller than any real number) could be considered a real number (and thus, the smallest real number), then as N approaches infinity, the range of each subset could be not 0 but 1 infinitesimal, with 1 infinitesimal/real number in each of an infinite number of subsets. In this case, the union of the _countable_ infinity of these subsets would be all real numbers between 0 and 1. Furthermore, there is no problem with any diagonal complement number for any of these subsets since every number n between 0 and 1 has a complement which is 1 - n that is contained in another of these subsets.3e:T852,## Extension of Cantor's Diagonalization Argument to Include all Permutations of the Enumeration It is often suggested that the diagonal complement number is not the only such number that supports Cantor's diagonalization argument. After all, the diagonal complement number is only one number that happens to be more easily identifiable, and if that were the only missing number then the enumeration would still be countable just by adding one. (i.e. infinity + 1 is still infinity) But, assuming the Reals are uncountable, there must be an uncountable number of other numbers not present in any countable enumeration of Reals. What other numbers might these be, and can we define them in terms of the enumeration itself, similar to how the diagonal complement number is defined? Extending Cantor's argument in various ways could allow us to find other similarly defined numbers that cannot be contained in the enumeration. Setting aside the question of whether the enumeration is complete, the enumeration is definitely countable, no matter what numbers it might be missing. So one thing we can do with this countable enumeration is that we can permute the order of the numbers resulting in a different order for the very same numbers in the enumeration. For example, we could swap the first two rows of the enumeration to result in a permutation that still contains the same numbers. Technically, since an enumeration is an ordered sequence, this new ordering represents a new enumeration, but since it is derived from the numbers in the original enumeration by merely permuting its members, let's refer to this as a **permutation of the enumeration**. When we permute the elements of an enumeration, the resulting permutation will have its own diagonal and diagonal complement numbers. The new diagonal complement number must not be present in the permuted enumeration, and it must also be absent from the original enumeration. Similarly, the original diagonal complement number must not be present in the permuted enumeration because the permuted enumeration contains the same numbers as the original enumeration. 3f:T6a2,There is an important property of any reasonable enumeration of real numbers represented in binary that we need to point out. If we look in every column of digits in the enumeration, we should expect to see an infinite number of rows with a '0' and an infinite number of rows with a '1'. If not, then the enumeration would clearly be missing infinitely many numbers that have '0's or '1's in that column. And furthermore, every consecutive span of '0's in each column will be finite in length, and the same is true for every consecutive span of '1's. If this were not the case, then an infinitely long span of '0's would mean that there could only be a finite number of '1's, and similarly, an infinitely long span of '1's would mean there could only be a finite number of '0's. Therefore, this property of having an infinite number of rows with both '0's and '1's and finite spans of consecutive '0's and '1's, holds for any reasonable enumeration of the real numbers. And consequently, this means that we can always find either a '0' or '1' in any column starting from any row by looking through only a finite number of subsequent rows. Back to the problem of creating a permutation such that a specific target number appears along the diagonal, because we can always find a row in the enumeration that contains the next digit of the target number by looking through a finite number of rows, and because this process can be repeated for each digit of the target number, the overall process of generating the entire target number is countable (that is, it has a countable infinity of steps). And that means it will always be possible to construct _any_ target number along the diagonal. 40:Ta07,### Generating any Number Along the Diagonal Results in a Contradiction Since we can build a permutation with any target number along the diagonal, let's consider what happens when the target number is the complement of an existing number in the enumeration. In this case, the existing number would be incompatible with the diagonal at every digit, so it will always be swapped with another number further down in the enumeration as the permutation is generated. Consequently, it will _never_ be used in a row of the permutation. To successfully permute an enumeration and generate a specific target number along the diagonal, we must ensure that every number in the enumeration is eventually used. However, this requirement will not be met if the diagonal number is the complement of an existing number in the enumeration, because that number can never be used in the permutation. So what would happen if we reject any such a number from the enumeration when is the complement of the diagonal? Let's start by generating the diagonal that is the complement of the first number in the enumeration. Then if we reject the first number, the remaining numbers in the enumeration will have exactly the same problem. We can still generate _any_ number along the diagonal using only the remaining numbers since we will have the same situation as described above for the original enumeration. It will still be true that we can find a number in the enumeration that is compatible with every digit of the diagonal by looking only a finite number of rows ahead. In fact, if we repeat this process any finite number of times, no matter how many numbers we have rejected from the enumeration so far, we can still generate _any_ number along the diagonal. By induction we can show that any finite number of numbers will be rejected from the enumeration, and we can continue on to infinity using transfinite induction, leading to the conclusion that _every_ number will be rejected from the enumeration. So it looks like we have a contradiction. Since we can permute an enumeration such that _any_ number can be generated along the diagonal, including any number that is in the enumeration, this seems to imply that the enumeration could not contain any numbers! But of course, we started with a countable infinity of numbers, so something is wrong. Is the problem with how we permuted the enumeration? Could it be that not every number can be generated along the diagonal? Or could it be that the definition of the diagonal complement number just doesn't make sense? 41:T88e,42:T4f7,43:T605,Perhaps a stronger argument is that, since we can permute the enumeration to generate any arbitrary number along the diagonal, and the complement of that number will be excluded, then every number will be excluded from the enumeration by some permutation. That is, if we collect all the excluded numbers for all the permutations of the enumeration, the result will be the same set as all the numbers in the enumeration, if not more. This is true even though, for each permutation of a permutation, there may be many other numbers that were in the original enumeration but are excluded by the permutation. These permutations would not be complete permutations, but subset permutations. Does it matter that a subset permutation does not include all the numbers in the enumeration? With an enumeration of a countable infinity of numbers, we can take just a subset that is also a countable infinity of numbers, and still generate a complete diagonal number. The general argument here is that, if it is legitimate to exclude the diagonal complement number from the enumeration because we define it to be a number generated from the diagonal digits of the enumeration that cannot be in the enumeration, then it should also be legitimate to show how we can exclude any number using some permutation of the enumeration, or even a subset of the enumeration. But that would imply that all numbers must be excluded with no numbers remaining in the enumeration, which is a contradiction since we started with a countable enumeration of numbers. 44:T795, But is it legitimate to make this argument based on subsets of the enumeration? Since a proper subset has fewer elements than the set it is derived from, that would exclude even more elements. In that sense, it is OK to show that we can exclude any arbitrary number even with an incomplete enumeration. But each enumeration that contains different numbers may theoretically result in a different set of diagonals. On the other hand, it appears that any sufficiently complex enumeration, with an infinite number of '0's and '1's in every column, can be used to generate any number along the diagonal. So this fact at least argues that it doesn't matter how incomplete an enumeration is. If we use the same starting enumeration each time we permute it to generate another diagonal number and thus exclude another diagonal complement number, this would avoid any problem with incrementally excluding more and more numbers, and then wondering whether we can still keep excluding all the rest. We can't show that a number is definitely in the enumeration unless we construct it specifically to have a certain value, but we can do that with our number generator system. If we construct an explicit number, even if it is a random number, but do not add it to the enumeration but instead, ensure that each diagonal digit is in conflict with this number in the corresponding digit, then we have shown that an arbitrary number, which could be any number, cannot be in the enumeration. To show that every number is excluded by some permutation, we have to show how a permutation of the enumeration can be generated such that the diagonal digits correspond to the complement of any arbitrary number. And since a permutation of the enumeration should have the same numbers as the enumeration, and we are showing that, for any number, there is a permutation that excludes, then all numbers must be excluded given all possible permutations.45:T8d5,Note that some permutations might result in a countably infinite subset of the enumeration. This would happen any time a permutation skips some number of elements of the enumeration, even a countably infinite number of elements. So another way to look at the generation of any number along the diagonal by selecting an appropriate permutation of the enumeration is that we don't even need to use all the numbers in the enumeration to generate a permutation that has a countable infinity of numbers such that the diagonal complement cannot be in that subset of the enumeration. We can ensure that only a subset of the enumeration that is countably infinite is used by _skipping_ numbers until we find the next number that has the desired digit in the next column. There will always be infinitely more numbers further down the list that have both '0' or '1' in every column. As long as the enumeration is not sorted such that all numbers with a '0' for a particular column precede all the rest of the numbers, and the same for '1's, which we can do, then there will always be only a finite number of numbers that we need to skip to find the next number. So what does this mean? There are always an infinite number of countably infinite subsets of the countable infinity of numbers in the enumeration, and for every subset, as long as we don't select numbers that have only one digit ('0' or '1') in a column, and don't order the numbers such that all the '0's or '1's come first, or last, then we can generate any number along the diagonal by permuting this subset or by creating yet another subset that has the target numbers along the diagonal. There is no bottom to this process. The subsets and subsets of subsets will be increasingly sparse by excluding more and more numbers, but we can still generate any desired number along the diagonal. Should there be a bottom beyond which the enumeration is too sparse? I don't think so. If a permutation skips some elements of the enumeration, the diagonal would be different from what results if all elements are included. But we don't need to skip any elements from the complete permutation if, after the first countable infinity of elements, we append the remaining elements not used by the first part.46:Tc1e,### Enumeration of Permutations of a Countable Infinity of Numbers Do we have to somehow enumerate all the permutations of the enumeration in order to make this argument about excluding all numbers from the enumeration, assuming that the diagonal complement number cannot be in any enumeration? That would be a problem if the number of permutations is uncountable. But I don't believe we do need to enumerate all the permutations to make that argument. We only need to say that for an arbitrary number, we can generate one permutation that results in the diagonal complement number being that number. This is a relatively simple refutation in that case. But given that we are arguing that Cantor's Diagonalization argument is invalid, and that the Reals may be countable after all, then the number of permutations of a countable enumeration of the Reals should also be countable. A generator of all permutations of a countable enumeration would follow a similar pattern as the generator of all Reals, but the problem is more complex since every element of a permutation can be mapped from any of the infinite elements of the enumeration. This is actually more closely related to an enumeration of all continued fractions, since each element of a continued fraction can have any finite value without limit. We will generate successively longer finite prefix permutations, but we have to also incrementally increase the maximum value for each element as the prefix grows longer. So, For the first step, we have 1 element, max of 1. ``` [[1]] ``` For the second step, we have 2 elements, max of 2. But we never repeat a number in a sequence, so [1, 1] and [2, 2] are not allowed (unlike how continued fractions which don't have this restriction). ``` [[1, 2], [2, 1]] ``` etc. These are only the finite length prefix permutations for an infinite sequence of all remaining values. For each of these finite prefixes, we need to randomly generate all the remaining values using a zig-zag method. We will incrementally increase the maximum random value the further out we go, starting with i+1 or maybe 2*i for position i. We also have to allow the numbers in the finite prefix to be included in the infinite random tail, but when we get to step n, the first n numbers will be any of the numbers from 1 to n, so that allows any of the numbers to eventually be in any position. So we are ensuring that every permutation of the first n numbers is generated this way, but how are we ensuring that every random infinite tail is generated? And how do we ensure that there are no possible duplicates such that the process would loop infinitely trying to find a unique permutation? This seems possible, but trickier than for the generation of numbers. The enumeration of all continued fractions will be easier than the enumeration of all permutations of the Naturals since, for the former, there is no condition on using each number only once as there is for the latter. But if we can do the former, then we can do the latter just by adding the condition that each number is used only once.47:T4ee,#### Diagonal and Diagonal Complement Numbers are Always Incompatible with a Balanced Enumeration Note that, if we are trying to generate any specific number along the diagonal that is already in a balanced enumeration by permuting the rows of the enumeration, that implies that the complement of that target number, which is also in the balanced enumeration, can never be used, so it will be forever pushed further down the list. And conversely, if we try to generate the complement of that number along the diagonal then its complement, the first number, can never be used. But if we are *not* trying to generate any specific number that is in a balanced enumeration along the diagonal, there is no problem with the number and its complement being in the enumeration. Nevertheless, the diagonal and the diagonal complement numbers will end up being *some* number in every permutation of the balanced enumeration, and no matter what they are, none of those numbers can be in the balanced enumeration. In fact, even without generating any permutations of the balanced enumeration, we know that *every* diagonal and diagonal complement number that could possibly result from *any* permutation will not be in the balanced enumeration, as we discussed before. 48:T446,### Enumeration of Other Subsets of Reals Cantor's Diagonalization argument is stated in terms of an enumeration of a subset of all Reals, all those between 0 and 1, which we know map 1-1 to all Reals. The argument is designed around the expectation that every diagonal complement number of any enumeration is also between 0 and 1, so it ought to be in the enumeration. Now that we know more about the nature of all possible diagonal and diagonal complement numbers for the original enumeration and the balanced enumeration that we can derive from it, we will investigate what happens with other subsets of the Reals besides the subset between 0 and 1. This argument *should* apply to enumerations of other subsets of the Reals, as long as we are careful to determine whether we should expect the diagonal complement number to be in the enumeration. In the following sections, we will attempt to enumerate the following subsets. 1. All real numbers between 0 and 1/2. 2. All irrational numbers between 0 and 1. 4. All rational numbers between 0 and 1. 5. All mirrored natural numbers. 49:T5ae,### Enumeration of Reals between 0 and 1/2 Should Also be Incomplete Suppose we instead provide a candidate enumeration of the set of real numbers between 0 and 1/2, excluding 1/2. While this set is the same size as the set of all reals, for every number in the enumeration of this set, its complement will be between 1/2 and 1, so it will not be included in this enumeration. And while every diagonal number resulting from a permutation of the enumeration will also be between 0 and 1/2, thus every diagonal *complement* number will be between 1/2 and 1, so they are all expected to be absent from the enumeration. Even though the diagonal complement numbers will necessarily be absent from the enumeration due to their construction, they are also _ not expected_ to be present in this enumeration. Therefore, Cantor's Diagonalization argument can't help us decide whether this enumeration is incomplete. But it is worth noting that Cantor's argument applies, in a general sense, to this subset of the Reals as well as to the subset between 0 and 1, though the fact that we don't expect any diagonal complement number to be in the enumeration means we can't conclude that the enumeration must be incomplete, and therefore we can't conclude that the set of Reals between 0 and 1/2 is uncountable. That doesn't mean that we know the set of Reals between 0 and 1/2 is countable, just that we can't conclude it is uncountable from Cantor's argument. 4a:T846,### Enumeration of Reals between 0 and 1/2 is Completed by its Complement Enumeration between 1/2 and 1 Finally, as argued above, every diagonal complement number constructed from _any_ enumeration between 0 and 1/2 is a number that is between 1/2 and 1, and vice versa. And if we map all the elements of one enumeration into its complement in the other enumeration, we can be certain that every diagonal complement number of one enumeration should be found in the other enumeration, and vice versa. The two enumerations are complements of each other. Instead of mapping from one enumeration to its complement, we could alternatively construct both enumerations simultaneously. For every digit added to a number in one enumeration, we will add its complement digit to the corresponding number in the complement enumeration. The result will be two complementary enumerations. Regarding the question of whether the diagonal number will necessarily be in the enumeration, we can ensure that it is definitely included in an enumeration without conflict by explicitly adding that number at a known location in the enumeration, say as the second number. The function that generates each of its digits will not generate digits randomly but use the diagonal digit of each subsequent number in the enumeration once it is added. If we know where we located the diagonal number, then we can also be certain that the complement of that number will definitely be at the same location in the complement of the enumeration. If we want to do the same with all the diagonal numbers that result from permutations of the enumeration, rather than explicitly generating each of the diagonal numbers, we will have to look them up in the permutation. We can be sure that we can _find_ each diagonal number somewhere in the enumeration since all combinations of digits will be generated. This look up process will take countably infinite time, but we can be sure the number will be found. And once we find the diagonal number, we then know that the corresponding diagonal complement number will be in the complement enumeration.4b:T8d5,### A Contradiction? If we merge these two enumerations (from 0 to 1/2, and from 1/2 to 1) into one enumeration, then we should have the same set of numbers as the enumeration from 0 to 1, and it will be a balanced enumeration since every number will be accompanied by its complement. This new enumeration will have its own diagonal and diagonal complement numbers, for each of the permutations of this enumeration. But because this is a balanced enumeration, that means that every diagonal number and diagonal complement number must be missing from the enumeration. But we can also construct the complement enumeration for this balanced enumeration, such that for each number in one enumeration, its complement will be in the complement enumeration. However, in this case, the complement of a balanced enumeration is the same enumeration where each number is swapped with its complement, so they are merely permutations of each other. We won't get anything new by complementing this enumeration since it contains its own complement. Also, every diagonal number and diagonal complement number will also be missing from this complement, which is consistent. However, whereas we could construct the diagonal number as one of the numbers in the enumeration of the reals between 0 and 1, we cannot do so here for an enumeration which should merely be a permutation of the same enumeration. Is this correct? Maybe there is something wrong with these enumerations. We could repeat the generation of the enumeration of reals between 0 and 1 but this time, we will simultaneously generate three other enumerations, one for the complement of every number between 0 and 1, one only for numbers between 0 and 1/2, and the other only for numbers between 1/2 and 1. For every number we generate in the first enumeration, we will also generate its complement and add that to the complement enumeration, and also add each of those two numbers to the other two enumerations depending on its value. 4c:T517,If we take the original enumeration and split it into two enumerations for numbers in [0, 1/2) and [1/2, 1], then the diagonal complements of the [0, 1/2) enumeration are in the complement enumeration from [1/2, 1], and vice versa. If we merge these two enumerations, the result will be a balanced enumeration that is identical to the enumeration from [0, 1]. The complement enumeration from [0, 1] should contain all the diagonal complements of the original enumeration, and vice versa. The merging of these two enumerations will be a balanced enumeration that suddenly can not contain any of the diagonal or diagonal complement numbers of the balanced enumeration, although it contains all the diagonal and diagonal complement numbers from the original and complement enumerations. But all three enumerations should be identical if they are complete. So this argument seems to suggest that the enumerations cannot be identical, and cannot be complete. If the original enumeration is complete, then it would be balanced and therefore the diagonal number should not be allowed in the enumeration because the diagonal complement number is not allowed. But clearly we can explicitly add the diagonal number to the enumeration, even though the diagonal complement number appears to be forced out. 4d:T8b5,#### More and more conflicts If we generate the enumeration of numbers between 0 and 1/2, there is no conflict with any diagonal complement since they would all end up in the complement enumeration from 1/2 to 1. So once we have that enumeration, we can multiply every element by 2 and result in the enumeration from 0 to 1. But suddenly we have more conflicts with the diagonal complement. Does this make sense? Or were there more conflicts in the enumeration from 0 to 1 that we hadn't considered? Do we have to consider a diagonal that starts in the second column, thus ignoring the first column? And if so, then we would have to repeat this for every diagonal starting at every digit after a finite length prefix. Yes, there was no guarantee that the diagonal complements that we discovered previously covered all of the conflicts. The general rule here is that, for every finite length prefix, if we collect all the numbers that have the same value for this same prefix, then the collection of all suffixes of all these numbers should be all the Reals, so this collection has the same number of conflicting diagonal complements as for all of the Reals. But we can generalize further. We could divide out the infinite collection of every other digit of each number that has the same value for those digits, and the remaining digits for all those numbers would be equivalent to all the Reals. Or in the most general case, we could take each and every combination of digit positions, divide the digits of each combination out from every number that has the same value for those digits, and the infinite string or remaining digits of all those numbers would be equivalent to all the Reals. How many combinations is this? It's the same as the number of Reals. I wonder if there is a way to eliminate all numbers this way. This would be similar to showing that we can permute the rows of an enumeration to generate any number along the diagonal, such that the diagonal complement cannot be in the enumeration, but that's a lot simpler than this recursive elimination of diagonal complements for all permutations of remainders after removing every subset of digits for all numbers that have the same value. lol 4e:T6a2,### Enumeration of Irrationals Should be Incomplete Assuming Cantor's Diagonalization argument is valid for the set of Reals between 0 and 1, the same argument might apply to candidate enumerations of the set of irrational numbers. We know that every diagonal complement number for any enumeration of the irrationals will necessarily be absent from the enumeration. However, there is a possibility that, while many of the diagonal complement numbers will be irrational, some of the diagonal complement numbers for some permutations could be rational numbers, in which case, we would *not* expect them to be in the enumeration of irrationals. But since the rationals are already known to be countable, we don't need to be concerned about their absence from the enumeration, and we can qualify the argument by noting that the diagonal complement is either rational, which we can ignore, or it is an irrational number that ought to be in the enumeration but is necessarily absent. Therefore it is safe to conclude that if Cantor's argument is valid for the Reals between 0 and 1, it is also valid, with a small qualification, for the Irrationals between 0 and 1. The question remains whether at least one diagonal complement for the enumeration is an irrational number. If it turns out that *every* diagonal complement is a rational number, then we don't know that the set of irrationals is uncountable, by Cantor's argument. Inverting the argument, if the irrational numbers are countable and if the enumeration is complete, then the diagonal and diagonal complement numbers must not include any irrational numbers so they must all be rational numbers. But that appears unlikely at best. 4f:Ta32,### Enumeration of Rationals between 0 and 1 Should be Complete Cantor's Diagonalization argument assumes that, given any candidate enumeration that is intended to be a complete enumeration of an infinite set in which the diagonal complement number ought to be included in that enumeration, if we find that the diagonal complement number is absent from the enumeration, then the enumeration must be incomplete, and therefore the infinite set must be uncountably infinite. This argument was used to justify the claim that the set of Reals between 0 and 1 must be uncountable. But suppose we leave out some numbers from the candidate enumeration, and we try to pass it off as a complete enumeration of the Reals. Suppose we leave out all of the irrational numbers so that the only numbers that remain are all rational. We can still apply Cantor's argument to this candidate enumeration, as long as we can prove that at least one diagonal complement number ought to be in the enumeration but it cannot be. Why would Cantor's argument not apply to this enumeration just as it does for one that is intended to include all real numbers between 0 and 1, or all irrational numbers between 0 and 1? One might respond that Cantor was only trying to show that the set of Reals cannot be enumerated, but there is nothing in the argument that requires that the enumeration must be an honest attempt at enumerating all Reals. To the contrary, since we know that the diagonal complement is necessarily absent from any enumeration, then it is necessarily absent regardless what set of numbers the enumeration actually does include, or what set of numbers it is intended to enumerate. Furthermore, if the diagonal complement of any enumeration is also expected to be in the set that is being enumerated in order for it to be considered a complete enumeration, then the absence of the diagonal complement number from any enumeration would prove that the set it is attempting to enumerate cannot be enumerated regardless what the set contains, assuming Cantor's argument is valid. And finally, if the set cannot be enumerated, then Cantor's argument concludes that the set must be uncountable. If you agree with the previous two sections that argue why Cantor's argument applies to the set of Reals between 0 and 1/2, though it doesn't help decide the countability of that set, and it applies to the set of Irrationals between 0 and 1, with a qualification about ignoring the diagonal complement numbers that are rational numbers, then why can we not also consider the infinite set of Rationals between 0 and 1? 50:T791,#### Enumeration of Rationals between 0 and 1 Cannot include Any Diagonal Complement Again, we know that every diagonal complement number that we construct from every permutation of *any* enumeration of any set of numbers will necessarily be absent from every permutation of the enumeration. This much we agree is true for any candidate enumeration of the Reals. And it is also true for our candidate enumeration that happens to be missing all irrational numbers. But before we can declare, by Cantor's argument, that the set associated with our candidate enumeration must be uncountable, we have to also prove that at least one of the diagonal complement numbers *ought* to be in the enumeration. We know that every diagonal complement number will be absent, but if at least one must be in the enumeration in order for it to be complete, then Cantor's argument allows us to declare that the enumeration must be incomplete. For any one enumeration of the set of Rationals between 0 and 1, the diagonal complement number might be an irrational number, or it might be possible for it to be a rational number. The set of all diagonal complement numbers for every permutation of the enumeration will include many irrational numbers, but it could also include some rational numbers. And if we can prove that some of the diagonal complement numbers will be rational numbers that *ought* to be in the enumeration, then we will have proved that the enumeration must be incomplete. And if this is true for *any* enumeration of some set, regardless what the set contains, then Cantor's argument would conclude that the set must be uncountable. So the key question in this refutation of Cantor's argument is whether the diagonal complement number for any enumeration of the set of Rationals between 0 and 1 could ever be a rational number between 0 and 1 that *ought* to be in that enumeration in order for it to be considered complete.51:T733,But any enumeration of the rational numbers will also be balanced, which would mean that, if the enumeration is complete, every diagonal and diagonal complement number for any permutation of the enumeration could not be in the enumeration, so it could not be a rational, so it *must* be irrational. We know it is possible to enumerate the rational numbers, but is it possible to create a permutation of a valid enumeration such that the diagonal number *must* be a rational number? If so, then we have a contradiction. Here is how we can generate a permutation of the enumeration of rational numbers between 0 and 1 such that the diagonal is a specific rational number. Starting with the first digit of the number, we will find a row of the enumeration in which the number in its first column has the digit we are looking for. That row will be swapped with the first row. Then we repeat the process with the second row and second digit of the number we are constructing along the diagonal, and so on for all remaining rows and columns. Since every column of the enumeration has an infinite number of 0s and 1s, and there is always a finite span of 0s between each pair of 1s, and a finite span of 1s between each pair of 0s, then we can always find a row within a finite distance from the current row that has the required next digit. (I forget whether I proved this is true for the Rationals, though it is clearly true for the Irrationals.) Thus we *can* construct any number along the diagonal. But will all numbers in the enumeration be used? One number that will be forced out of the enumeration between 0 and 1, including 1, is the diagonal complement number, since it will necessarily be different from every number that is in the enumeration. And if the diagonal number is rational, so is the diagonal complement number.52:T74f,However, note that the limit set of the sequence of numbers in the enumeration will include this diagonal complement number, since we can find a sequence of numbers that is a subset of the enumeration such that each number has an increasingly longer prefix of digits that is identical to the digits of the same length prefix of the diagonal complement number. Each of these numbers will have a prefix length that is shorter than the row index of the number, because otherwise the diagonal digit will always be in conflict. So we can conclude that the limit set of the sequence of numbers in the enumeration will include the diagonal complement number. But can we use a separate line of reasoning to also conclude that the limit set cannot include the diagonal complement number? Every increasingly longer number will be different from the diagonal complement number in at least one digit, the diagonal digit, so the sequence cannot include the diagonal complement number. So it would seem that, at the limit, the diagonal complement number could not be in the limit set. This is another way to look at the paradox of Cantor's Diagonalization argument. But forcing the diagonal digits to be a specific sequence is not the same as finding out what the diagonal digits end up being while not forcing them one way or another. For an enumeration of *all* the rational numbers between 0 and 1, we would expect every rational number and its complement, also a rational, to be in the enumeration, and this would mean that the diagonal number could not be rational. But could it be an infinitesimal? If the enumeration just happened to be ordered such that all the diagonal digits are 0s, except for infinitely far out, we "finally" get to the diagonal complement number composed of all 1s, then the diagonal number would represent the first infinitesimal, 0.000...0001. 53:Ta49,### Enumeration of Mirrored Natural Numbers Should be Complete In order to see that any enumeration of the set of all Rationals between 0 and 1 might result in a diagonal complement number that is rational, we will consider an even "smaller" subset of the Rationals that is, nevertheless, a countably infinite set. This subset is the set of so-called "mirrored" natural numbers which are constructed by reversing the digits of each natural number and prepending the decimal point. These are rational numbers between 0 and 1 that are sums of a finite number of powers of single base, base 2 in this case. For example, 6 in base 2 is "110", so the mirrored version of this number in binary is "0.011", which is 1/4 + 1/8. You may have an objection to constructing the mirrored natural numbers this way because it will require natural numbers that are infinitely long in order to construct mirrored numbers with infinitely long decimals. But note that we are not trying to represent numbers with infinitely long decimals here since every number has a finite length. Equivalently, each number is the sum of a finite set of negative powers of 2. For example, the mirror image of the finite number 6 is "0.011" which is 1/4 + 1/8. These numbers will all have a finite number of '1' digits corresponding to the powers of 2 in the sum, and an infinite number of '0' digits appended to the end, since no other power of 2 will be included in the sum. It is also important to note that each rational number between 0 and 1 represented in binary that has a finite number of '1' digits and an infinite number of trailing '0' digits can also be equivalently represented by another number with a finite number of '0' digits and an infinite number of trailing '1' digits. For example, "0.011" is equivalent to "0.01011111..." with repeating '1's. This equivalent number is *not* the same as the complement, which would be "0.10011111...", and the complement is equivalent to _another_ number which is "0.10100000..." in this case. Do we need to include these equivalent numbers in the enumeration in order to be complete? Since every complement number is equivalent to some other number, we _could_ include every complement number, and consequently we will be including every equivalent number, but then we will have duplicate numbers in the enumeration. So we only need to include one of each pair of equivalent numbers. The first few rows of an enumeration of the mirrored natural numbers in binary, without the complements or equivalent numbers, ordered the same as the natural numbers they are derived from, will look like this:54:T15a7,(Posted this summary of the whole argument to a math forum, which initially received mostly civil response, but quickly collapsed into exaggerated misunderstandings, on their part, of what I actually said.) Imagine, if you will, that someone has created an enumeration of real numbers, not just an enumeration of rational numbers as is so often proposed. The numbers in this enumeration would have to be expressed as calculations which will provide as many digits of each number as requested, since of course we cannot write all the digits out. Furthermore, imagine that we have a convincing proof of the completeness of this enumeration, regardless of Cantor's proofs that the enumerations must be incomplete. Suspend your disbelief just for a moment, and imagine that this is what happened. Maybe you'll have to time-travel back to before Cantor, before limits, maybe even earlier, sometime after Newton and Leibniz created the calculus. Then, someone, maybe Cantor himself, dreams up the nested closed intervals proof and the diagonalization proof of the incompleteness of any possible enumeration. But since there is already a proof that an enumeration is possible, would the proofs of incompleteness still be accepted as valid, and more convincing than the previous proof of completeness? OK, enough drama.. What am I talking about? I do have an enumeration of real numbers between 0 and 1, one that is not simply an enumeration of rational numbers, but a generator of infinitely long sequences of all combinations of binary digits of real numbers created in such a way that completeness of the enumeration can be proved. No one, as far as I know, has come up with the enumeration that I am about to present, along with the proof of its completeness. And the existence of such an enumeration calls into question what is actually going on in Cantor's arguments. I am also not simply attacking Cantor's arguments from the outset, since admittedly, they do appear convincing, once you make the effort to fully understand them. Most of the counter-arguments I have seen are easily shot down, though each seems to have an inkling of some real underlying problems, often assumed to be merely misunderstandings or confusions. So here is an outline of the argument, broken down into 4 parts. 1. We can create a system of number objects to represent real numbers as strings of binary digits. For each number, only a finite number of digits will be known at any finite time, and a function is provided to compute as may of the remaining digits as requested. All digits of all numbers for a single enumeration will be stored in a single binary tree, and an enumeration table can also be displayed. See Number Objects for the details, and a few examples of using this system to enumerate subsets of Rational numbers. 2. Any single real number can be computed by randomly generating all of its digits. This is not to say that we know, at this point, how to enumerate all real numbers, but just that any single real number can be generated. Even this possibility requires questioning the current understanding about true randomness and the computability of numbers. And the question of whether the probability of generating a specific real number purely randomly is 0 or an infinitesimal drags in the Hyperreals. See Computable Real Numbers for this argument, which includes examples of computing a few specific real numbers, one random number at a time, and a countable infinity of random real numbers. This last enumeration can not be considered a complete enumeration, however, due to the possibility of leaving out some numbers and the possibility of looping infinitely while generating the same random number more than once. 3. An enumeration of all real numbers can be generated, however, if we are careful to avoid the gaps and the possibility of conflicts that could otherwise arise in the generation of random numbers. A breadth-first algorithm is presented that does that, though it may at first glance appear to be doing nothing more than the many previous attempts to fill out a binary tree with rational numbers. The key difference is that, for each node in the tree corresponding to a finite prefix of digits, we can associate one unique real number with its infinite suffix of all remaining digits. The completeness of generating every possible finite-length combination of digits is easy enough to see, but this argument also extends to infinite-length strings by using transfinite induction. Even so, the association of every real number (between 0 and 1) with one infinite-length combination of digits, one unique path through the tree, is another problem, which can also be shown by transfinite induction. See Countably Infinite Random Real Numbers for this argument. 4. Finally, having established at least the plausibility of a complete enumeration of the reals, we can look with a fresh perspective at Cantor's proofs than any such enumeration must be incomplete. His three proofs may all be variations of the same diagonalization type of argument, though the first two based on nested intervals, also explicitly involve limits that may be implicit in the third proof. But for all three, it appears a contradiction results if we apply the diagonalization not just once, but a countably infinite number of times to the same candidate enumeration, such that every possible number in the enumeration will be in conflict with the diagonal complement. This contradiction is presented in Refutation of Cantor's Diagonalization.55:[] 30:{"id":"5460bd8257b874be","slug":null,"title":"","thumbnail":"7959c836108161e53c99327b02fc260b66e88881aa3d61bb60a3ccd6d3568808","access_level":"public","create_time":"$D2022-12-22T22:43:35.074Z","update_time":"$D2023-01-15T16:33:04.002Z","view_time":null,"first_public_time":"$D2022-12-22T22:43:35.074Z","has_unpublished_edits":null,"fork":{"id":"2555290b866cbe32","slug":"refutation-of-cantors-diagonalization","title":"Refutation of Cantor's Diagonalization","owner":{"id":"88bca6b25e968fae","login":"dlaliberte","name":"Daniel LaLiberte"}},"stars":0,"starred":false,"owner":{"type":"team","id":"88bca6b25e968fae","login":"dlaliberte","name":"Daniel LaLiberte","avatar_url":"https://avatars-next.observableusercontent.com/e5ca1f0f640d7d3579ea69f7316220ebcec267322399e2773f31b91bc2790b15","tier":"starter_2024"},"version":1447,"publish_version":null,"role":null,"authored":false,"authors":[{"type":"individual","id":"b887ef310da1b17a","login":"dlaliberte","name":"Daniel LaLiberte","avatar_url":"https://avatars-next.observableusercontent.com/e5ca1f0f640d7d3579ea69f7316220ebcec267322399e2773f31b91bc2790b15","tier":null}],"members":[{"type":"individual","id":"b887ef310da1b17a","login":"dlaliberte","name":"Daniel LaLiberte","avatar_url":"https://avatars-next.observableusercontent.com/e5ca1f0f640d7d3579ea69f7316220ebcec267322399e2773f31b91bc2790b15","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"lapsed":null,"subscribed":null,"forks":1,"collection":null,"type":"notebook","body":{"title":"","cells":[{"id":767,"value":"Work in progress... This notebook is barely begun, starting with a clone of a prior version of a previous notebook, \n[Refutation of Cantor's Diagonalization](https://observablehq.com/@dlaliberte/refutation-of-cantors-diagonalization)\n\nIf you are interested in this subject, please read the earlier notebooks first:\n\n- [Number Objects](https://observablehq.com/@dlaliberte/number-objects)\n- [Computable Real Numbers](https://observablehq.com/@dlaliberte/computable-real-numbers)\n- [Countably Infinite Random Real Numbers](http://observablehq.com/@dlaliberte/countably-infinite-random-real-numbers)\n- [Refutation of Cantor's Diagonalization](https://observablehq.com/@dlaliberte/refutation-of-cantors-diagonalization)","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":0,"value":"# Refutation of Cantor's Nested Closed Intervals\n\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1187,"value":"## ...\n\n---\n\n---\n\nSome of the diagonalization argument is relevant for the Nested Intervals argument.\n\n---\n\n---\n\n---","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":775,"value":"## Abstract\n\nIn the previous post, [Countably Infinite Random Real Numbers](http://go/plx-fabric/@dlaliberte/countably-infinite-random-real-numbers), we showed how the real numbers could be enumerated using a breadth-first generation of random numbers, and we presented proof of its completeness. Here we examine Cantor's nested closed intervals argument regarding any candidate enumeration of the real numbers to see what happens with this enumeration. ","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":771,"value":"toc(\"h2, h3, h4, h5, h6\")","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":472,"value":"## Breadth-first Enumeration of Countably Infinite Reals\n\nLet's review the countable enumeration of Reals that was presented previously. We can see that the [Enumeration of Countably Infinite Reals Breadth-First](https://observablehq.com/@dlaliberte/countably-infinite-random-real-numbers#cell-218), copied below, generates real numbers with every combination of digits for every finite length prefix. Keep in mind that every real number generated by this function only _starts_ with a finite length prefix and is followed by an infinite number of random digits. These are not merely rational numbers.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":516,"value":"### A Frequent but Flawed Refutation\n\nMany people who attempt to refute Cantor's argument point out that, because the diagonal complement digits generated from the first `d` numbers _can_ appear in the enumeration, but further down in the enumeration after the first `d` numbers, that effectively refutes Cantor's argument. But that counter-argument reflects a misunderstanding of Cantor's argument. He would certainly agree that a finite-length prefix of the diagonal complement can appear in many numbers further down the list. Nevertheless, for every one of those numbers, its diagonal digit will conflict with the corresponding digit of the diagonal complement number.\n\nBut furthermore, Cantor was not arguing about just a finite-length prefix of the diagonal complement number but the entire infinite length of this number generated by complementing the diagonal digits of every number in the enumeration.\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1185,"value":"$32","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":898,"value":"#### Nested Intervals around the Diagonal Complement\n\nWe can see how the diagonal complement number is excluded from the enumeration by using a variation of the generator of one random real number. We start by adding a pre-generated random real number as the first number in the enumeration, and we also use each of its digits as the diagonal digits of every number. Furthermore, every number that we add to the enumeration will have a prefix that is the complement of this diagonal number up to, but not including, the diagonal digit.\n\nThis is the same visualization we showed earlier, but here we add nested interval brackets around the diagonal complement prefixes, closing in on the diagonal complement number that will never be added.\n\nIn the enumeration table below, you can see that, while the diagonal number can be in the enumeration, the diagonal complement number can never be in the enumeration, at least not at any finite position.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":907,"value":"mutable nRandomDigits = []","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":905,"value":"function generateNRandomDigits(n) {\n // Slice off extras\n if (nRandomDigits.length > n) {\n nRandomDigits.splice(n);\n }\n // Push on new digits.\n while (nRandomDigits.length < n) {\n nRandomDigits.push(Math.random() < 0.5 ? \"0\" : \"1\");\n // db(\"nd\", nRandomDigits);\n }\n return nRandomDigits;\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":900,"value":"$33","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":902,"value":"diagonalComplementNumbers = generateOneRandomNumberAsDiagonal(\n 0 + diagonalComplementOptions.depth,\n diagonalComplementOptions.pregenerateX,\n false\n)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":910,"value":"html`\n${button({\n value: \"Regenerate\",\n divStyle: {\n display: \"inline-block\"\n },\n buttonStyle: {\n background: \"#CCE9FF\"\n },\n titleStyle: {\n \"text-align\": \"center\"\n },\n onclick: ({ button }) => {\n mutable nRandomDigits = [];\n }\n})}`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":914,"value":"viewof diagonalComplementOptions = form(html`\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":917,"value":"vlembed(\n makeTreeVegaSpec(diagonalComplementNumbers.tree, {\n title: \"Create X along Diagonal\",\n width: 600 + diagonalComplementOptions.depth * 5,\n height: Math.max(300, 30 + diagonalComplementOptions.depth * 12),\n hideLabels: diagonalComplementOptions.depth > 60,\n linkShape: \"diagonal\", // diagonalComplementOptions.linkShape\n colorScheme: \"bluepurple\"\n })\n)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":921,"value":"showEnumerationTable(diagonalComplementNumbers)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1329,"value":"// viewof db = Debug()","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1414,"value":"$34","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1271,"value":"$35","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1274,"value":"breadthFirstIntervals = generateIntervalsBreadthFirst(\n Math.pow(2, 1 + breadthFirstIntervalsOptions.depth)\n)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1276,"value":"//viewof replayGenerateIntervalsBreadthFirst = html``\nhtml`\n${button({\n value: \"Regenerate\",\n divStyle: {\n display: \"inline-block\"\n },\n buttonStyle: {\n background: \"#CCE9FF\"\n },\n titleStyle: {\n \"text-align\": \"center\"\n },\n onclick: ({ button }) => {\n mutable nRandomDigits = [];\n }\n})}`","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1279,"value":"viewof breadthFirstIntervalsOptions = form(html`\n`)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1282,"value":"vlembed(\n makeTreeVegaSpec(breadthFirstIntervals.tree, {\n title:\n \"Generate Countable Reals with Intervals Breadth-First to Depth \" +\n (breadthFirstIntervalsOptions.depth + 1),\n width: 600 + breadthFirstIntervalsOptions.depth * 10,\n height: Math.max(\n 300,\n 50 + Math.min(2000, Math.pow(2, breadthFirstIntervalsOptions.depth) * 18)\n ),\n hideLabels: breadthFirstIntervalsOptions.depth > 7\n })\n)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1290,"value":"showEnumerationTable(breadthFirstIntervals)","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1434,"value":"$36","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":531,"value":"$37","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":45,"value":"### The Sequence of Nested Intervals and the Limit of the Sequence are Defined Self-referentially\n\nFirst note that the sequence of nested closed intervals is defined self-referentially. Each subsequent interval must be nested within the previous interval. Also, the limit of the sequence is defined in terms of the sequence of intervals. \n\nThe limit of a sequence of closed intervals is thought to be a single number. But the innermost nested interval always remains an interval, not a single number, even at the limit. ","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1230,"value":"$38","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1252,"value":"### Nested Intervals with Unidentified Limit\n\nWe are required to create nested closed intervals that don't contain any of the previous numbers in the enumeration. But a possible problem with constructing these intervals is that the closed intervals themselves must be bounded by numbers as well, so which numbers are we going to use for the interval bounds, while simultaneously excluding numbers from the intervals?\n\nHowever, that construction should be equivalent to the following variation: Starting from an interval for the entire range, 0 to 1, for each number that is created, if it is contained within the most deeply nested known interval, we will add a new interval by replacing either end of the previous interval with the new number. Then we repeat this process indefinitely. The sequence of intervals converges to a number we have not yet identified. As soon as we find another number that is inside the innermost interval, we use that number as one of the bounds of the next deeper interval.\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1254,"value":"$39","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1256,"value":"$3a","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1258,"value":"$3b","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1260,"value":"Finally, there remains a problem with the definition of the limit of a sequence of nested closed intervals which precludes the possibility that the limit of the difference between the two innermost bounds will always be a non-zero infinitesimal. This suggests that the definition of limits is flawed, and that the real numbers are not complete unless we admit the infinitesimals, and their inverses, the infinities.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1262,"value":"$3c","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1053,"value":"$3d","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1441,"value":"---\n\n---\n\nThe rest is copied from earlier versions of the refutation of diagonalization. It probably doesn't reflect all of my current thinking, and it is not relevant to the nested intervals argument except for the common diagonalization argument. Leaving it here for now since I have to leave it somewhere...\n\n---\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":73,"value":"$3e","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":720,"value":"\nIt's worth noting that we don't need to physically permute the enumeration to determine its resulting diagonal complement number. Instead, we could simply permute the row indices of the numbers in the enumeration to generate the diagonal complement number for the permuted enumeration. \n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":81,"value":"### Every Diagonal Complement for Any Permutation *Must* be Absent from Every Permutation\n\nWe can apply this argument to every possible permutation of the enumeration. For each permutation, there will be a corresponding diagonal complement number that must not be present in any of the permutations of the same enumeration. Since some permutations may have the same diagonal complement numbers as other permutations, a natural question to ask is how many unique diagonal complement numbers will be generated for all possible permutations?\n\nIt is worth emphasizing that none of the diagonal complement numbers for any of the permutations can appear in any of the other permutations. This is because the diagonal complement for each permutation will necessarily be different from every number in that permutation and in _every_ permutation of the enumeration, since all permutations contain the same numbers as the original enumeration.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":819,"value":"If we want to generate a list of all possible permutations of the original enumeration of the Reals, we might have a problem, since the number of such permutations grows even faster than the number of subsets of a set. Specifically, the number of possible permutations of a set of size `n` is `n!`, while the number of elements in the power set (the set of all subsets) of a set of size `n` is ${tex`2 ^ n`}. \n\nThe set of natural numbers is a countable set, but the power set of the natural numbers, which is the set of all subsets of the natural numbers, is assumed to be an uncountable set. Therefore, the set of all permutations of the natural numbers (or any countable set) would also be uncountable, because it would contain even more elements than the power set. And this would apply to our set of real numbers as well.\n\nSo for now, we will not be able to enumerate all possible permutations of the enumeration of reals, even if the enumeration is complete and thus the set of Reals is countable.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":758,"value":"Regardless of the size of the set of permutations, we can nevertheless define it mathematically and derive from it another set containing the unique diagonal complements from each permutation. Every one of these diagonal complement numbers must be absent from all permutations of the enumeration.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1090,"value":"### We Can Choose a Permutation that Results in Any Specific Diagonal Number\n\nNow let's turn the problem around to ask: Can we select the appropriate rows of the enumeration to generate the corresponding permutation such that a specific target number appears along the diagonal? \n\nTo make this possible, for each digit of the target number, we need to be able to find a previously unused number in the enumeration. We should also try to use every number in the enumeration eventually, but this requirement turns out to be central to the problem posed by Cantor's Diagonalization argument.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1106,"value":"$3f","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":135,"value":"\nTo build a permutation of an enumeration that generates a specific target number along the diagonal, we will compare each digit of the target number with the diagonal digit of each subsequent row. If the diagonal digit matches the target number, we can move on to the next digit and the next row. If it does not match, we need to find another row further down to swap with this one before we can move on.\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":546,"value":"$40","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1123,"value":"### The Cause of the Contradiction\n\nWe can re-examine each step more closely, but the most suspicious component is the self-referential contradictory definition of the diagonal complement number. We should conclude that this definition does not correspond to a valid number, and thus there is no conflict with any number in the enumeration and no number in the set of reals between 0 and 1 has been identified as necessarily missing from the enumeration by using this construction.\n\nBut note that, even if we have successfully refuted Cantor's diagonalization argument, that doesn't mean we have proved that the Reals are countable. It only proves that Cantor's proof that they are uncountable is invalid. But we nevertheless have separately proved the completeness of the breadth-first enumeration of the reals, which constitutes proof that all Reals are countable.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":1115,"value":"## End of Document\n\nThe remainder of this document is perhaps extraneous at this point. Leaving it here for now to mine for other good ideas. You can see how long and tortuous the process has been for me.\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":116,"value":"","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":182,"value":"$41","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1095,"value":"$42","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":542,"value":"But what happens if we try to generate a permutation of the enumeration such that the diagonal number is one of these diagonal complement numbers? Is that even possible? Yes, since we can generate any specific diagonal number, as shown above! Note that the number that corresponds to the diagonal digits may or may not be in the enumeration, and if the diagonal digits of this permutation correspond to one of the previous diagonal complement numbers, then this diagonal complement number is no longer in conflict with the diagonal digits, so it _could_ appear in the enumeration. But alas, the complement of the diagonal digits in this permutation correspond to a new diagonal complement number that cannot be in the enumeration, even though it was previously in the enumeration.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":549,"value":"If we weren't trying to generate any specific diagonal number, would the generation of all permutations of the rows of one enumeration include the permutation that results in one of the diagonal complement numbers? Yes it would, assuming we could enumerate all these permutations, which we will get to later.\n\nThus, it seems that we have a paradox at least. Cantor's Diagonalization argument claims that the diagonal complement number cannot appear in the enumeration, and yet we can generate a permutation of the enumeration that results in the digits of that diagonal complement number appearing on the diagonal, except that pushes out the previous diagonal number.\n\nTo say that more precisely, if an enumeration results in diagonal d and diagonal complement c, then we can permute the enumeration to result in a new diagonal that is the same as c, which forces out the new diagonal complement which is the same as d.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":637,"value":"$43","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":726,"value":"$44","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":722,"value":"$45","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":733,"value":"Should we change the idea of an enumeration to be any countable enumeration, even one that is transfinite? Is there even a way to restrict an enumeration one that is not transfinite? Is this lack of transfinite enumerations the reason that diagonal complements are forced out of the first countable non-transfinite enumeration? If we go this route, then it may be considered valid and non-contradictory to define the diagonal complement number as it is defined, since it is allowed for the entire enumeration to transcend beyond the first countable infinity.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":640,"value":"$46","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":239,"value":"$47","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":266,"value":"$48","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":292,"value":"$49","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":552,"value":"If we merge these two enumerations (from 0 to 1/2, and from 1/2 to 1) into one enumeration, then we should have the same set of numbers as the enumeration from 0 to 1, and it will be a balanced enumeration. One way we can tell that we have the same number of members in all three enumerations is that we can simply map each of the numbers in the enumeration from 0 to 1/2 into a number in the enumeration from 0 to 1 by multiplying by 2, and we can similarly add -1 to each number to result in the complement numbers in the enumeration from 1/2 to 1. And we can do the reverse mapping. So we have a one-to-one and onto mapping, thus a bijection between these sets.\n\nBut if Cantor's Diagonalization argument cannot be used to determine that the enumeration between 0 and 1/2 is incomplete, why should it apply to the enumeration between 0 and 1, where both enumerations have the same number of elements and we can map from one to the other?","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":618,"value":"$4a","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":622,"value":"$4b","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":626,"value":"$4c","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":633,"value":"$4d","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":264,"value":"$4e","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":282,"value":"$4f","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":392,"value":"$50","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":412,"value":"$51","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":565,"value":"$52","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":315,"value":"$53","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":348,"value":"Table(countingNumbers.map(formatMirroredNatural), tableOptions)","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":332,"value":"countingNumbers = [\n 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":346,"value":"tableOptions = ({\n layout: 'fixed', \n sort: 'index',\n align: {\n index: 'right',\n \"approximate value\": 'left'\n }, \n width: {\n \"index\": 50,\n name: 50,\n digits: 500,\n \"approximate value\": 100\n },\n format: {\n index: (n) => `#${n} | `,\n \"approximate value\": (n) => n.toString(), \n digits: (d) => html`${d}`\n }\n});","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":343,"value":"function formatMirroredNatural(num) {\n const boldDigits = { \"0\": \"[𝟎]\", \"1\": \"[𝟏]\" };\n const normalDigits = { \"0\": \" 0 \", \"1\": \" 1 \" }; // Note: using 'punctuation space ' '\n function emboldenDiagonal(digits, numIndex) {\n return digits\n .split(\"\")\n .map((digit, digitIndex) => {\n return numIndex == digitIndex ? boldDigits[digit] : normalDigits[digit];\n // const diagonalDigit = boldDigits[digits[index]];\n // digits = digits.substring(0, index) + diagonalDigit + digits.substring(index + 1);\n })\n .join(\"\");\n // return digits;\n }\n const pattern = \"0̅\";\n const binaryDigits = (num) => num.toString(2).split('').reverse().join('').padEnd(16, 0);\n return {\n index: num, // .index,\n digits: `0.${emboldenDiagonal(binaryDigits(num), num)} ${pattern}`\n };\n}","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":422,"value":"If we create a balanced enumeration, by adding the complement of each number in the enumeration inserted after the number it complements, then every other row will have a '1' digit in the diagonal, so the resulting diagonal will be \"0.010101...\", and the diagonal complement will be \"0.101010...\". This is a rational number that is not expressible as a sum of a finite number of negative powers of 2, since it requires an infinite series of negative powers of 2. This is consistent with the diagonals and diagonal complement numbers all being absent from the balanced enumeration. \n\nFor some permutations of this enumeration, the diagonal and diagonal complement number will be rational numbers that are not expressible as a sum of a finite number of negative powers of 2, and the rest will be irrational. In no case will the diagonal or diagonal complement numbers be one of the numbers in the enumeration.","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":452,"value":"(1 / 8).toString(3)","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":395,"value":"\nStill haven't done this, and it may not be possible...\n\nIn other words, if Cantor's Diagonalization argument is valid for the Reals, then it should be just as valid for the Rationals, with the same kind of qualification that we added regarding the set of Irrationals. But we know by an independent proof, also by Cantor, that the Rationals are actually countable, so that must mean that Cantor's Diagonalization argument isn't valid for arbitrary enumerations in which diagonal complement numbers that we expect to be in the enumeration cannot be.\n","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1445,"value":"$54","pinned":false,"hidden":false,"mode":"md","author":"$undefined","summary":"$undefined","output":"","database":"$undefined"},{"id":502,"value":"import { toc } from \"@nebrius/indented-toc\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":329,"value":"import { Table } from \"@observablehq/table\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":487,"value":"import { form } from \"@mbostock/form-input\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":498,"value":"vlembed = require(\"vega-embed\")","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":500,"value":"import {\n Real,\n Rational,\n Zero,\n One,\n BinaryTreeNode,\n makeTreeVegaSpec,\n showEnumerationTable,\n digitsToNumber\n} from \"@dlaliberte/number-objects\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":912,"value":"import { button } from \"@bartok32/diy-inputs\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1009,"value":"import { Debug } from \"@trebor/debug-tool\"","pinned":false,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"},{"id":1013,"value":"// viewof db = Debug();","pinned":true,"hidden":false,"mode":"ojs","author":"$undefined","summary":"$undefined","output":"$undefined","database":"$undefined"}],"files":"$W55","stdlib":"1","titling":"auto"}} 31:"$30:body"