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Cantor's proof represented the set of real numbers as a matrix like so...

$1 \to 0. d_{1,1} d_{1,2} d_{1,3} \dots$

$2 \to 0. d_{2,1} d_{2,2} d_{2,3} \dots$

$3 \to 0. d_{3,1} d_{3,2} d_{3,3} \dots$

$\dots$

where each $d_{x,y}$ is a binary digit $0 \lor 1$

then the proof created a real number from the diagonal by flipping a bit in each row

$diagonal = 0. \sim d_{1,1} \sim d_{2,2} \sim d_{3,3} \dots$

This $diagonal$ could not be in row $1$ or $2$ or $3$ or any row $n$ where $n \in \mathbb N$

But, we can look at this from a different perspective.

If we let row $2^n = 0. \sim d_{1,1} \sim d_{2,2} \sim d_{3,3} \dots \sim d_{n,n} r_1r_2r_3 \dots$

where the first $n$ digits is the first $n$ digits of the diagonal flipped. Followed by random bits $r_1r_2r_3 \dots$ then for every value $n$ where $n \in \mathbb N$ we can rule out our diagonal number being in rows $1$ to $n$ but we can't rule out row $2^n$ as the first $n$ digits of this row is the same as the first $n$ digits of the diagonal.

My question is, what rigorous method allowed Cantor's proof to use that version of the diagonal proof, rather than this inconclusive version of the proof?

My own thoughts on this problem is, of course we can rule row $2^n$, because the $2^n$th digit of the diagonal can't be the same as the $2^n$th digit of row $2^n$ but it can't rule out row $2^{2^n}$ or $4^n$. If we rule out $4^n$, we can't rule out $8^n$. If we rule out $8^n$, we can't rule out $16^n$ and so on forever.

As we look further and further along the diagonal, we will be getting closer to looking at an infinite number of digits in the diagonal $(n \to \infty)$ which will be the same as the same near infinite number of digits in our row $2^{(n \to \infty)}$. So as we approach infinity, the two numbers seem to converge, and if they do converge, then the impossible diagonal could be in the list, and Cantor's proof would fail.

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    $\begingroup$ Possible duplicate of How does Cantor's diagonal argument work? $\endgroup$ – Mark S. Jun 1 '18 at 14:57
  • $\begingroup$ Not that it really matters here, but note that $2^{2^n}$ is not the same as $4^n$. For example, $2^{2^3}=256$ but $4^3=64$. $\endgroup$ – Henning Makholm Jun 1 '18 at 15:05
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So as we approach infinity, the two numbers seem to converge, and if they do converge, then the impossible diagonal could be in the list, and Cantor's proof would fail.

This appears to be the crux of your argument: By choosing the starting list cleverly you can make it so that there is a subsequence of the list that converges to the diagonal number.

That much is true. (In fact, if you choose a list that contains every rational number at least once, or even just every number with a finite decimal representation -- which is easily done -- then every real number will be the limit of some subsequence of the list, no matter whether it's the diagonal number or not).

However, the fact that there is a subsequence of the list that converges to the diagonal number does not mean that the diagonal number itself is in the list. And all the diagonal construction claims is there is some real number that is not in the list.

At this point you may want to claim something like "... but the diagonal number will be at position $2^{\infty}$ in the list!" That won't work, however, because by definition the list is a map from $\mathbb N$ to $\mathbb R$, and there is no such natural number as $2^{\infty}$, so whatever that means, it is not in the domain of the map.

(If you allow extending the domain of the function from $\mathbb N$ to something larger, you can of course have a bijection $(\mathbb N\cup X)\to\mathbb R$ for some appropriately chosen $X$. This does not contradict Cantor's result).

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    $\begingroup$ +1. To the OP: another way of phrasing the paragraph beginning with "However" is: "The antidiagonal sequence may be in the closure of the list (in a sense which can be made precise), but that's not the same as being literally on the list." The reals are in fact the closure of a countable set in a precise sense, but that's not what the diagonal argument is arguing against. $\endgroup$ – Noah Schweber Jun 1 '18 at 15:09
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You are comparing the diagonal to every mapping from a natural number to a real number. It does not match with the number mapped by 1 because the first digit is different. It does not matter if a later number has the same first digit as our diagonal. We only care that the first digit is not a match with the mapped number to the number 1.

Then, you compare with the number mapped from 2 and find that it differs on at least the second digit. So, it is not a match with that one.

You continue down to each natural number and find that the diagonal number does not match any number that is mapped.

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If we let row $2^n = 0. \sim d_{1,1} \sim d_{2,2} \sim d_{3,3} \dots \sim d_{n,n} r_1r_2r_3 \dots$

As the prover, you don't get to “let” row $2^n$ be anything. You are supposing for the sake of contradiction that the real numbers can be arranged in a sequence. The entire sequence is provided, and once the diagonal number is constructed, you can't go back and rearrange the sequence.

Maybe you're asking “what if”, for each $n$, row $2^n$ agreed with the diagonal number in each of the first $n$ positions. That would be very coincidental, would not represent a logical gap in the proof. The diagonal number, by construction, disagrees with the number in row $2^n$ at the $2^n$th position. So the diagonal number is not in row $2^n$.

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