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Very sorry in advance if these are naive questions, but I am still having a bit of trouble digesting Rudin's Diagonalization argument in the proof of Theorem 2.12 in Principles of Mathematical Analysis.

So far I have read the following posts (the first three give the proof as in the text verbatim, I would've copied it here again but I'm not sure how. Very sorry!) :

This one, this one, this one, and this one.

The third of which most closely matches my question, but I'm still slightly confused.

  1. (EDIT: This was explained to me in the comments) How is Rudin's argument not circular? At some point, wouldn't the diagonal will be a sequence of countably many elements? In fact, wouldn't there be countably many diagonals which map to countably many elements of $\mathbb{N}$?

If this is true, then the full diagonal sequence he constructs would then be the union of these countably many diagonals with countably many elements, and then claiming the full diagonal sequence is countable would be what he is wanting to prove?

  1. Is there anything special about Diagonals (I suspect not, given some of the alternative proofs given in these overflow answers, but just to clarify). For example could the sequence be constructed by "tiling" the matrix using fixed sized sub-matrices?

    By this I mean counting the elements within an $n \times n$ submatrix, for each such submatrix in the following way:

For example, with $n =2$ starting from the top right corner in the following pattern:

[(0,0), (0,1), (1,0), (1,1)], [(0,2), (0,3), (1,2), (1,3)], [(2,0), (2,1), (3,0), (3,1)], [(2,2), (2,3), (3,2), (3,3)]...

and continuing this pattern, counting the elements within the matrices from the top left corner to the bottom right (This is the pattern I was thinking, but I guess this is still moving along the diagonal... but maybe this could be done differently?)

For the $i$-th sub-matrix you enumerate, you map it to the natural numbers $\{(i-1)n^2+1, (i-1)n^2+2, \dots, (i)n^2 )\}$.

Shy
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    I am not sure if this is what you're asking, but diagonals are indeed special (a better word is perhaps helpful) in mathematics. For example in topology there is a very useful result that says a space $X$ is Hausdorff if and only the diagonal in $X \times X$ is closed. Another example is what's known as a "diagonal sequence trick", which is a commonly used trick in analysis to extract a convergent subsequence with certain desired properties from a larger sequence. – CBBAM Jul 26 '23 at 19:53
  • Oh, ok interesting! Thank you very much for this insight! Would you happen to be able to offer any intuition as to why they are so useful as opposed to other patterns of iterating? – Shy Jul 26 '23 at 20:00
  • "At some point, the diagonal will be a sequence of countably many elements." -- This is not true. Each and every diagonal is of finite length. – PrincessEev Jul 26 '23 at 20:06
  • "How is this not circular argument?" -- There is no circular argument, because he is not presuming the result. He shows the existence of a bijection $\mathbb{N} \to S$, by rearranging the elements into a sequence. – PrincessEev Jul 26 '23 at 20:06
  • @PrincessEev Sorry I'm a bit confused still, so the length of the $n$-th diagonal is $n$, and even though you are essentially "sending $n$ to infinity" it is still finite for each instance because it's fixed for each instance? Even though $n$ is infinite in the limit? – Shy Jul 26 '23 at 20:14
  • "For example could the sequence be constructed by "tiling" the matrix using fixed sized sub-matrices?" Can you clarify what you mean by this? Lots of your question suffers from lack of clarity. You need to be more explicit in each part of it. Also, Can you explain why you think "Rudin's argument is circular"? – Adam Rubinson Jul 26 '23 at 20:34
  • @AdamRubinson Thanks for your patience, and feedback. This is my first post so I will try to clarify what I am asking and do better in the future! – Shy Jul 26 '23 at 21:16
  • " it is still finite for each instance because it's fixed for each instance? Even though n is infinite in the limit?" -- Perhaps to rephrase matters: which of the diagonals is infinite in length? If you claim one to exist, surely you can find an example of an infinite diagonal. The diagonals may grow in length without bound, but each and every diagonal is finite in length despite that. – PrincessEev Jul 26 '23 at 21:32
  • @PrincessEev I see! That helped a lot, thanks very much! This incorrect assumption was at the heart of my misunderstanding that the argument was circular, so this makes sense now! (Also I'm not sure if it's possible for me to mark your answers as helpful, but they were!) – Shy Jul 26 '23 at 21:35

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Perhaps the double indexing in the matrix construction is what is confusing you, if this is the case I recommend you think of this in a different way.

You know (I presume) that $\mathbb{N}^2$ is countable. (It is fairly easy to explicitly construct a bijection between $\mathbb{N}$ and $\mathbb{N}^2$)

But $\mathbb{N}^2$ is a countable union of countable sequences, the first sequence could be

$ ((0,0),(0,1),(0,2),...)$

With the next sequence being

$((1,0),(1,1),(1,2),...)$

And so on. The proof you are questioning is simply generalising this to arbitrary countable sequences, but any countable sequence can be mapped to surjectively to from a sequence of the form

$ ((n,0),(n,1),(n,2),...)$

So we can surjectively map $\mathbb{N}^2$ to the countable union of countable sequences, and composing this with the bijection from $\mathbb{N}$ to $\mathbb{N}^2$ we get a surjection from $\mathbb{N}$ to the countable union of countable sequences, and an injection between these two is easy to create, so there must exist a bijection.

So the only relevance of the "diagonals" really is that they are used most frequently to construct the bijection between $\mathbb{N}^2$ and $\mathbb{N}$, and they are relatively nice in this regard since this bijection has a sortof polynomial form (I say sortof since it has $2$ arguments, not sure if there's a better term for it) You could tile it like you were asking, you could do it in all kinds of chaotic ways, but I doubt any of them would have a comparably nice description

Carlyle
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