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Can a set be infinite, countable, and compact? Can you please review my proof attempt and guide me?

Note: This question is asking about my proof attempt. My goal is to learn, and you only learn by trying to prove things. Thus, I ask that this question be reopened. I've edited this post to make this more clear.

I tried to prove that it cannot (at least in $\mathbb{R}$), but my proof attempt contains an error.

Proof Attempt:

If set $S \subset \mathbb{R}$ is infinite and countable, then $S$ is not compact.

Let $E: \mathbb{N} \to S$ be an enumeration of $S$. Let $w(n) = \min (|E(n+1) - E(n)|, |E(n-1) - E(n)|)$, with $w(1) = |E(2) - E(1)|.$ Let $O(n)$ be the open interval $(E(n) - w(n)/3, E(n) + w(n)/3)$. Then every $s \in S$ is in exactly one $O_n$ [*], and so the collection $O = {O_n : n \in N}$ is an open cover of $S$ that does not admit a finite subcover. QED.

[*] This is not true, since the enumeration is not in any order. To fix it, we'd need $w(n)$ to be $\inf \{|E(m) - E(n)| : m \in \mathbb{N}\}$, but that is equal to zero.

Questions:

  1. Can a set be infinite, countable, and compact?
  2. Is there any way to salvage my proof attempt? Perhaps to prove something weaker.
  3. If so: How far can it be extended? To $\mathbb{R}^n$? Beyond?
SRobertJames
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    Take any convergent sequence on any metric space. – Andrew Nov 11 '22 at 19:16
  • Interestingly, a topological group cannot be infinite, countable and compact. Even more interestingly, there is a one line proof of this involving measure theory and functional analysis! – FShrike Nov 11 '22 at 19:18
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    @AndrewZhang Take any injective, convergent sequence in any metric space, and union in the limit. – Theo Bendit Nov 11 '22 at 19:20
  • @FShrike Can you please elaborate? I found https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality but cannot follow it. – SRobertJames Nov 11 '22 at 19:28
  • @TheoBendit Can you please elaborate? Or perhaps expand what you are saying into an answer? – SRobertJames Nov 11 '22 at 19:29
  • @SRobertJames The MO post uses complicated ideas freely. The proof I alluded to is the top-voted answer invoking the Haar measure, whose existence requires nontrivial functional analysis and measure theory. Given the existence of the Haar measure, this allows for a deceptively simple proof of the result. Yemon's and Mariano's topological arguments are also quite nice, though I don't know the result about infinite profinite groups being uncountable. To explain these results you'd need to know the contents of a decent introductory topology course, I'd say. If you want, ask this as a question – FShrike Nov 11 '22 at 19:37
  • However, the question would have to be made precise. After all, most of the results they use are available online for you to research yourself. If the research is too hard to read (this happens all the time!) then asking on MSE about it is suitable – FShrike Nov 11 '22 at 19:38
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    Just to clarify what the commenters are saying: The statement you are trying to prove is false, and one can obtain a counterexample from any convergent sequence together with its limit point. And so no, it is not possible to salvage your proof. General advice: try not to prove false statements. – Lee Mosher Nov 11 '22 at 19:38
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    @LeeMosher I actually think it's a very good exercise to try to prove false statements. That is the best way to build intuition and understand why and where they are truly false. Note that Leslie Lamport relates that he discovered the Paxos algorithm when he was trying to prove that any algorithm with such properties could not exist: Attempting to prove a false statement led Lamport to one of the greatest discoveries of theoretical (and applied) computer science. – SRobertJames Nov 11 '22 at 19:46

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