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Given the set $B$ of all possible sequences of binary digits of finite length ("finite binary sequences"), construct a finite binary sequence $S$ such that each member of the sequence is the logical negation of the corresponding submember of the corresponding member of $B$, or because that sounds a bit ridiculous: $$\textrm{let}\ S = \{ S_n \ | \ S_n = \neg{B_{n_n}} \forall n \in [1, 2, 3,...] \}$$

$$B_1 = {\color{blue}{1}, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,...} \\ B_2 = {0, \color{blue}{1}, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0,...} \\ B_3 = {1, 0, \color{blue}{0}, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0,...} \\ B_4 = {1, 0, 0, \color{blue}{1}, 1, 0, 0, 1, 0, 0, 1, 0, 1,...} \\ B_5 = {1, 1, 0, 0, \color{blue}{1}, 1, 1, 0, 1, 1, 1, 1, 0,...} \\ B_6 = {0, 1, 1, 1, 1, \color{blue}{0}, 0, 1, 0, 1, 0, 1, 0,...} \\ B_7 = {1, 1, 0, 0, 0, 0, \color{blue}{1}, 0, 1, 0, 1, 0, 1,...} \\ B_8 = {1, 1, 1, 0, 1, 0, 1, \color{blue}{0}, 0, 0, 1, 1, 0,...} \\ B_9 = {1, 0, 0, 0, 0, 0, 0, 1, \color{blue}{1}, 1, 0, 1, 0,...} \\ B_{10} = {0, 1, 1, 0, 0, 0, 0, 1, 0, \color{blue}{1}, 0, 0, 1,...} \\ B_{11} = {0, 0, 1, 0, 0, 1, 0, 1, 1, 0, \color{blue}{1}, 0, 0,...} \\ B_{12} = {1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, \color{blue}{1}, 1,...} \\ B_{13} = {0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, \color{blue}{0},...} \\ \vdots \\ ------------------- \\ \ \ \ S = {\color{red}{0}, \color{red}{0}, \color{red}{1}, \color{red}{0}, \color{red}{0}, \color{red}{1}, \color{red}{0}, \color{red}{1}, \color{red}{0}, \color{red}{0}, \color{red}{0}, \color{red}{0}, \color{red}{1},...}$$

Because each member $S_n$ of $S$ is defined as the logical negation of the corresponding value in the corresponding member of $B$, $S$ cannot be equal to any member of $B$. $S$ differs from $B_1$ at position $1$, $B_2$ at position $2$, $B_3$ at position $3$, and so on. So $S$ is not in $B$. And we've got a contradiction.

We defined $B$ as the set of possible finite binary sequences, and yet we created a set $S$ that is both a finite binary sequence and not in $B$, so where's the contradiction? Considering I never tried to construct $B$ but was simply provided it as the set of all possible binary sequences, am I to conclude that it is simply impossible for a set to contain all possible binary sequences? Or that it is impossible to evaluate membership of such a set? Or perhaps that because the sequences are by definition of finite length, there simply aren't enough digits to accommodate negating all possible members of $B$?

I'm a bit confused, because while I understand the theory behind the use of this diagonal argument in proof, there's a logical contradiction where it seems all math is sound. Things get even more odd when the requirement for the sequences to be of finite length is dropped and $B$ is made to be the set of all possible binary sequences. What am I missing?

TheEnvironmentalist
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    Why should $S$, defined this way, be finite? – G Tony Jacobs Jul 18 '17 at 01:11
  • Do you have a typo somewhere in this sentence? We defined B as the set of possible finite binary sequences, and yet we created a set S that is both a finite binary sequence and not in B, so where's the contradiction? –  Jul 18 '17 at 01:13
  • Since we have infinite many lines, $S$ must be an infinite sequence. – Peter Jul 18 '17 at 01:15
  • @Peter I thought of that, and it gets tricky right there. There are infinitely many sequences of finite length, in that you could never find the last one. And yet in the same grain, you could never find the last digit of the last member of $B$. So the number of digits matches the number of elements such that it works out, as far as I can tell. But if this is where the contradiction is, please let me know – TheEnvironmentalist Jul 18 '17 at 01:38
  • S most certainly is not finite. However if you were given every sequence of infinitely countable sequences you would get a contradiction. And the resolution of that would be that there is no countable list of all countable seqences and the nmber of infinite sequence is not countable. Here the set of finite sequences is countable but if you made an finite list of all seqences you would get a contradiction that is resolved in that the set of all finite sequences is not finite. – fleablood Jul 18 '17 at 01:39
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    What is "the last member of $B$"? – Noah Schweber Jul 18 '17 at 01:44
  • @NoahSchweber haha that one really pained me to write out. The point being that the length of $S$ is limited by the length of the elements of $B$, so as far as I can reason length shouldn't be a factor limiting its membership in $B$, as every member came from an equally-large member of $B$ to the length up to that index, ad infinitum – TheEnvironmentalist Jul 18 '17 at 01:47
  • "The point being that the length of S is limited by the length of the elements of B" which are completely unlimited. – fleablood Jul 18 '17 at 01:49
  • @TheEnvironmentalist No, its length is absolutely a factor limiting its membership in $B$. $B$ is the set of finite sequences. $S$ is not finite, any more than the set of natural numbers is finite (even though each natural number itself is finite). Do you understand this? – Noah Schweber Jul 18 '17 at 01:49
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    Instead of listing all sequences lets list some sequences. And lets list them so that $B_1$ has length 1. And let $B_2$ have length $2$ and $B_k$ has lenghth $k$. So how long are the longest members? – fleablood Jul 18 '17 at 01:52
  • @TheEnvironmentalist Note that this is exactly the same difficulty you had in this earlier question; you are having trouble distinguishing actually-finite things from infinite objects which are "built out of" finite pieces. – Noah Schweber Jul 18 '17 at 02:40
  • @NoahSchweber It would indeed seem so. To help me understand, answer me this: If $S$ gets its $n^{th}$ member from the $n^{th}$ member of $B_n$, and we can agree that $S$ is infinite in length, then $S$ is clearly longer than nearly every member of $B$. And yet $S$ can't be longer than every member of $B$, because it needs sources for its members. I have no problem seeing that $S$ is longer than infinitely many members of $B$, but I can't see how any closed set of axioms allows $S$ to be strictly larger than every member of $B$ when it draws its length directly from that of $B$'s members. – TheEnvironmentalist Jul 18 '17 at 03:36
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    @TheEnvironmentalist "And yet $S$ can't be longer than every member of $B$, because it needs sources for its members." That's nonsense. $S$ can indeed be longer than every element in $B$, because there's no single element of $B$ that $S$ relies on. $S$ gets to draw on all the elements of $B$, and $B$ has arbitrarily long elements. – Noah Schweber Jul 18 '17 at 03:41
  • @NoahSchweber I'll play the other side for a minute. Let's take this at face value and agree that $S$ is indeed strictly longer than every member of $B$. Because of what strictly longer means, and because its length is discrete, and as such can change by values no smaller than 1, it must be at least 1 larger than every element of $B$. And yet if it is at least 1 larger than every element of $B$, then it has a member that is not present at member $n$ in any member of $B$. This is a contradiction, as its members all come from the matching indices of $B_n$, a member of $B$. – TheEnvironmentalist Jul 18 '17 at 03:56
  • @TheEnvironmentalist No, that's not right. You're conflating "For each element $f$ of $B$, there is some initial segment of $S$ of length greater than that of $f$" with "There is some initial segment of $S$ of length greater than that of every element of $B$." That is, you're conflating "$\forall \exists$" with "$\exists \forall$". This is not a thing you can do. For instance, the statement "For every number $x$, there is a number $y$ such that $x<y$" is true, but the statement "There is a number $y$ such that for every number $x$, $x<y$" is false. This is a fundamental error. – Noah Schweber Jul 18 '17 at 04:03
  • @NoahSchweber I really appreciate this discussion by the way. I'm learning a lot of exactly what I wanted to learn, so I'm really thankful you're taking the time to help. Here's my current perspective, if you wouldn't mind please break it somewhere: Either $S$ is strictly larger than all elements of $B$, or it isn't, because, well, there is no third option between is and isn't. Now if $S$ isn't strictly larger than all elements of $B$, then because the inverse of strictly larger than all is no larger than some the size of $S$ is bounded by some elements of $B$.... – TheEnvironmentalist Jul 18 '17 at 04:38
  • ... Given that all elements of $B$ are finite, so then is $S$. Alternatively, $S$ is strictly larger than all members of $B$, in which case, because larger in the context of the discrete number of elements in a set means has more elements, $S$ has more elements than all members $B_n$ of $B$, so there exists some $S_n$ for which $B_{n_n}$ does not exist, which defies the definition of $S$. Either $S$ is strictly larger, leading to a logical contradiction, or $S$ is not strictly larger, leading to a logical contradiction. – TheEnvironmentalist Jul 18 '17 at 04:38
  • @TheEnvironmentalist "$S$ has more elements than all members BnBn of BB, so there exists some $S_n$ for which $B_n$ does not exist" No, that's not true. Again, you're mixing up the logic: just because no single $B_n$ provides all the digits of $S$, does not mean that there's some digit of $S$ which is inaccessible from every element of $B$. This is exactly the same mistake as before: you need to realize that "For every $x$, there is some $y$ such that [stuff]" is not the same as "There is some $y$ such that, for every $x$, [stuff]." – Noah Schweber Jul 18 '17 at 04:42
  • You are falling into a very common stumbling block. For each one of the many $S_n$ there will be a $nB$ that is length equal or longer than $n$. From that you are making a mistake that therefore the most be some single ${\alpha}B$ that is longer than all $S_n$. This is not true because the $nB$ that are each longer than some $S_n$ are different and there is not one ${\alpha}B$ that works for all. This is EXACTLY the same as: for every $n$ there is an $m > n$ therefore there is some $m$ larger than every number. So all numbers are less than $m$ and $m$ is the largest number. – fleablood Jul 18 '17 at 19:24
  • "S has more elements than all members $B_n$ of B, so there exists some $S_n$ for which $B_{n_n}$ does not exist". Why do you believe that. That simply not true. For any $S_n$ there is a $B_{m_n}$ that is longer than $S_n$. But so what? There is also an $S_{p_{m_n}}$ longer than $B_{m_n}$. And $S$ is longer than all of these. – fleablood Jul 18 '17 at 19:40
  • Infinite sets and intuition don't mix well. Consider the set of all natural nunbers up to a maximum m, N(m). Every member of the set is finite, and its cardinality is finite. Similarly, every member of the set of all finite numbers N is finite, but its cardinality is infinite. This seems like a contradiction to many, but it is a fundamental property you need to grasp before you can understand Cantor. In this question, B can only be constructed if you know that len(Sn)<=n, or you pad each with infinite zeros. Either way, S has that same unintuitive property, that it's length has to be infinite. – JeffJo Jul 19 '17 at 15:51
  • @JeffJo Am I wrong in assuming that this behavior is largely arbitrary, i.e. we could have made infinities behave in a few different ways, but because we chose this particular path, all the resulting mathematics functions in this way? – TheEnvironmentalist Jul 20 '17 at 03:07
  • I'm not sure what you mean, but I think the answer is "no". We can make infinite sets "behave" in many different was that would be contradictory if they were finite, and it happens with all of them. You can match the set of all natural numbers 1:1 with the even natural numbers by the relation e=2n. In fact, Cantor defines an infinite set as one that can be put in a 1:1 with a subset of itself, so there is nothing arbitrary about it. Your S has to be infinite because it has to have a character in position n, for every n in N. – JeffJo Jul 20 '17 at 10:55

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We defined $B$ as the set of possible finite binary sequences, and yet we created a set $S$ that is both a finite binary sequence and not in $B$, so where's the contradiction?

The contradiction is exactly what you said: You constructed an $S$ that by definition of $B$ should be in $B$ but is not in $B$.

Did you perhaps mean to say $S$ is an infinite binary sequence? In that case, yes, there is no contradiction. In fact, the set of all finite binary sequences is indeed countable.

I believe the confusion you're having is because you're applying Cantor's diagonal argument to the wrong thing. The original diagonal argument was applied to the set of all infinite binary sequences, not the set of all finite binary sequences. So your definition of $B$ must be adjusted.

  • I'm not trying to duplicate Cantor directly, I'm asking a reduced version, and one that, to the best I can tell, is a perfectly valid argument to make regardless of whether applied to the finite or to the infinite – TheEnvironmentalist Jul 18 '17 at 04:41
  • @TheEnvironmentalist What do you mean by "reduced version"? It isn't valid for finite because (1) as you pointed out (assuming you did in fact mean to say that $S$ is infinite, which it is) there is no contradiction and (2) the set $B$ as you defined is not uncountable, so no proof that it is uncountable will be valid. –  Jul 18 '17 at 11:16
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and yet we created a set $S$ that is both a finite binary sequence and not in $B$,

Why do you think $S$ is finite? In fact, it's easy to show that it won't be - and this resolves your difficulty.

Noah Schweber
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  • The length of $S$ is, because of its member-by-member construction, no larger than the length of the longest elements in $B$, it qualifies as a member of $B$ just as much as any other member of $B$, no? – TheEnvironmentalist Jul 18 '17 at 01:41
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    There is not longest element in B. Just as there is no largest finite number. – fleablood Jul 18 '17 at 01:41
  • @TheEnvironmentalist And what do you think is "the length of the longest element in $B$"? Do you think there is a largest natural number? – Noah Schweber Jul 18 '17 at 01:44
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    The fact that the elements are themselves finite and bounded, does not mean the set of all of them is finite and bounded. There is no longest sequence as for any sequence of length k one can make a sequence of length k+1. As the list is infinitite the sequense S is infinite. – fleablood Jul 18 '17 at 01:44
  • "it qualifies as a member of B just as much as any other member of B, no?" No. It most certainly does not. – fleablood Jul 18 '17 at 01:45
  • @fleablood Hmm, I'd really appreciate if you expanded that logic to an answer so I can think it over and try to understand – TheEnvironmentalist Jul 18 '17 at 01:48
  • @TheEnvironmentalist Let's start small. Do you understand why the set of natural numbers is not finite? – Noah Schweber Jul 18 '17 at 01:49
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    Okay. Suppose S is finite. How long is it? Say it is $k$ long. But then what is $S_{k+1}$. If $S$ is finite there can not be any $S_{k+1}$ term. But you defined it so that $S_{k+1} = \lnot B_{k+1,k+1}$. So there is an $S_{k+1}$ term. S is not finite. For any $k$ there is a finite sequence that is $k$ long. ANd there are an infinite number of possible $k$s. There is no longest $k$. – fleablood Jul 18 '17 at 02:06
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    For any finite number $n$ there is another finite number $n+1$. So there is no largest natural number. And for any finite sequence that is $n$ long, there is another that is $n+1$ long. So there is no longest finite sequence. So S is not bounded by the longest sequences because there aren't any longest sequences. – fleablood Jul 18 '17 at 02:09
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One inconsistancy, that is !!!!!NOT!!!!! the error, is that you define $S_n$ as $\lnot B_{n,n}$. But as $B_n$ is finite, what is $S_n$ if $B_n$ has length shorter than $n$ (and as there are $2^k$ sequences that are length $k$ and $2^k > k$ there must be some $B_n$ that are shorter than length $n$[$*$]).

!!!!THIS IS NOT THE ERROR!!!!!!.

We can define $S_1 = \lnot B_{1,1}$. Then $S \ne B_1$. If $B_{2}$ through $B_{k}$ are less than $2$ long, don't do anything. Let $i_2$ be such that $B_{i_2}$ is the first sequence after $B_1$ that is larger than equal to $2$. Then let $S_2 = \lnot B_{i_2, 2}$. This way $S \ne B_1$. $S \ne B_{2}$ through $B_{i_2-1}$ (as those are all shorter than length $2$ and $S \ne B_{i_2}$ as $S_2 \ne B_{i_2, 2}$.

Continue so that $S_n = \lnot B_{i_n, n}$. Then $S$ will not equal any of the $B_i$ and each term is well defined.

But as every $S_n$ exist, that means $S$ is INFINITE. Your insistence that $S$ is finite is bizarre as for every possible $n$ an $S_n$ term exist so there are an infinite number of terms. So $S$ is CLEARLY infinite.

You confusion seems to lie in that every finite SUBsequence of $S$ from $S_1.....S_m$ is finite and is shorter than some $B_{i_m}$ that somehow $S$ the sequence that contains all the terms must be shorter than some $B_{i_m}$. That is simply wrong.

For any $S_1.... S_n$ there is some $B_m$ that is longer but there is NOT !one! $B_m$ that is longer than all $S_1 ..... S_k$. For each $S_1..... S_n$ there is a DIFFERENT $B_m$ that is longer. And there is NO $B_m$ that is longer than ALL $S_1...S_n$.

Which really should be obvious... So $S$ is very clearly and inarguably infinite.

fleablood
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  • I really like this answer, the conclusion being that there simply aren't enough values of $B_n$ to construct S. But that would mean that $S$ cannot be drawn from $B$. The problem I have is stating that $S$ can be drawn from $B$, and $S$ can be infinite while all $B_n$ are not – TheEnvironmentalist Jul 18 '17 at 03:45
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    But...no. that isn't the issue. The issue is simply that S is defined to have an infinite number of terms. So it isn't finite. – fleablood Jul 18 '17 at 04:19
  • Is an infinitely large sequence required to be strictly larger than any and all sequences of finite length? – TheEnvironmentalist Jul 18 '17 at 04:43
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    What does "larger than all" mean? Yes, it is longer than any. But "longer than all" doesn't have any meaning. – fleablood Jul 18 '17 at 19:11
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I sense you are not seeing the forest for the trees. Cantor's diagonal argument shows that the reals cannot be put in a list. The reals have the same cardinality as the power set of the integers. This shows the power set $P (\mathbb Z) $ has higher cardinality than $\mathbb Z $. The result generalizes to power sets of any set. It's easy to prove there is no surjection from a set to its power set. Given $f:X\rightarrow P (X) $, define y in P(X) by $y=\{x|x \notin f(x)\}.$ Then y is not in the image of f. No surjection means the power set has higher cardinality.

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    Would you mind explaining what "cannot be put in a list" means? I can understand that such a list cannot be constructed by any algorithm. That would be consistent with undecidability and Gödel incompleteness. I can accept that there's no way of making such a list, but how can they not be put in a list? {ℝ} there, that's the list. How doesn't that work? – TheEnvironmentalist Jul 18 '17 at 03:42
  • @TheEnvironmentalist given such a list of numbers, I can produce a number you have left out... by simply going along the diagonal and constructing a number which differs from the ith number in the list in the ith place after the decimal. .. –  Jul 18 '17 at 03:50
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    A list is broadly defined (and very intentionally so) as a collection of items, with or without an associated ordering. A collection of items is... just a collection of items. The pile of bananas on my counter is a collection of items. The keys on this keyboard are a collection of items. How are the reals not a collection of items? There's something profoundly nonsensical about the notation that a list, or set if you like, defined only as a collection of things, cannot contain a certain well-defined collection of things – TheEnvironmentalist Jul 18 '17 at 04:02
  • They're an uncountable collection. .. –  Jul 18 '17 at 04:09
  • That's vocabulary – TheEnvironmentalist Jul 18 '17 at 04:33
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Cantor's diagonal argument is almost always misrepresented, even by those who claim to understand it. This question get one point right - it is about binary strings, not real numbers. In fact, it was SPECIFICALLY INTENDED to NOT use real numbers.

But another thing that is misrepresented, is that it is a proof by contradiction. It isn't, and actually can't be. A proof by contradiction has to show that a contradiction follows from what was assumed, and it has to follow from ALL of what is assumed. Let me give a silly example. Assume the moon is made of a combination of bleu and green cheese, in a ratio that is equal to the square root of two. A well-known use of proof by contradiction is that if you assume a ratio is the square root of two, you get a contradiction about the factors in that ratio. This proves that there can't be a ratio equal to the square root of two. But it doesn't prove anything about the cheeses that make up the moon, even if you include that in your assumption. The contradiction does not follow from the assumed cheeses.

That the anti-diagonal string is not in your list does not follow from assuming you had a list of ALL strings. Only that you had a list of SOME strings. So all it proves is that any list of SOME strings can't be ALL. But that is enough: proving "if A, then B" also proves "if not B, then not A." This is called proof by controposition, not contradiction.

The basic outline of the proof is: 1) Assume T is the entire set you are interested in. 2) Assume S is a countable subset of T. 3) Create the anti-diagonal string of S, called D, which cannot be in S. 4) Prove that D is in T. 5) This proves "If S is Countable, it is not all of T." 6) By contraposition, "If S is all of T, it is not countable."

Where the argument in the original question fails, is step 4. The anti-diagonal it called S must be infinite since it is a function of the infinite set B. Therefore it isn't supposed to be in B.

JeffJo
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First understand that the proposed set $B$ is a countably infinite set of finite sets and is isomorphic to the rational numbers, and is possible to construct.

When you construct the diagonal $S$ across the entire list, that will produce another countably infinite set. So you are comparing apples to oranges. The list $B$ has finite sets and the diagonal $S$ is a countably infinite set.

If you still want to try to define a finite set $S$ out of the diagonal then that set will have a last element located at the end of some $B_X$ in the main set $B$. Later on at some at some point $B_Y$ the $S$ string will appear if the list is really complete.

Either way there is no contradiction involved.

user439222
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