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In Rudin's Chapter 2, Section 2.44, the Cantor set is described as a union of infinitely many intervals. It is true that we can find an infinite open cover that covers these intervals, but we cannot find a finite subcover within these infinite open covers to cover the Cantor set. For example, we can use an open cover to cover $[{(3k) \over 3^m },{(3k+1) \over 3^m }]$, but you cannot find a finite subcover within this open cover. This is because the Cantor set is composed of an infinite union of $[{(3k) \over 3^m },{(3k+1) \over 3^m }]$and $[{(3k+2) \over 3^m },{(3k+3) \over 3^m }]$. From this perspective, the Cantor set can be considered non-compact. Could someone point out any flaws in my argument above that the Cantor set is non-compact?

Update

Thanks for Eric's explanation. I have understood why P is compact, but now I have a new question.

$P=\bigcap\limits_{n=1}^{\infty} E_{n}$ essentially means that$\lim\limits_{n \to \infty}E_n$, where for a fixed positive integer n, $E_n$ contains intervals, right? However, as n tends to infinity, the lengths of those intervals will approach zero. Can I understand that these intervals then become individual points? And the distances between these points also tend to infinitesimal, so Rudin's proof goes as follows:

To show that P is perfect, it is enough to show that P contains no isolated point. Let$x \in p$, and let S be any segment containing x. Let $I_n$ be that interval of $E_n$ which contains x. Choose n large enough, so that $I_n \subset S$. Let $x_n$ be an endpoint of $I_n$ , such that $x_n\neq x$. It follows from the construction of P that$x \in p$. Hence x is a limit point of P, and P is perfect

Rudin pointed out that x is a limit point of P because as n approaches infinity, for any neighborhood of x, there exists a corresponding $x_n$ in the neighborhood of x. This is because, according to the construction of the set P, the endpoints of the intervals $I_n$ are not discarded. In other words, for any $x_n$ that belongs to $I_n$ and is an endpoint of $I_n$, $x_n$ also belongs to set P. Additionally, as the length of the interval $I_n$ tends to infinitesimal when n approaches infinity, the distance between x and $x_n$ also tends to infinitesimal. This implies that for any neighborhood of x, there will always be an $x_n$ in the sequence that belongs to set P and lies within the neighborhood of x. Therefore, x is referred to as a limit point of set P.

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  • $\begingroup$ The Cantor set is not the union of any family of (non-trivial) intervals, infinite or otherwise. Did you miss the part where it says "$P=\bigcap_{n=1}^\infty E_n$ is the Cantor set" ? $\endgroup$ Sep 2, 2023 at 9:20
  • $\begingroup$ Cantor set is not a union of intervals. It does not contain any (non-degnerate) interval. $\endgroup$ Sep 2, 2023 at 9:20
  • $\begingroup$ Each $E_n$ is a union of closed intervals, hence $E_n$ is closed, and thus $P=\cap E_n$ is also closed. The boundedness of $P$ follows from $P\subset [0,1]$. Any bounded closed set is compact. $\endgroup$
    – Feng
    Sep 2, 2023 at 9:23
  • $\begingroup$ Could you please review my new understanding? Thank you in advance $\endgroup$
    – ssds
    Sep 4, 2023 at 13:49
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    $\begingroup$ Good-natured comment: Because MSE is a question-and-answer site, not a math discussion site, it would be preferable to roll back the question to its original form, mark Eric's answer as accepted, and ask a new question about "the intervals [becoming] isolated points", linking back to this question as needed. $\endgroup$ Sep 4, 2023 at 15:07

1 Answer 1

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The cantor set is the intersection of countably many closed sets and is therefore closed. It is bounded and thus by Heine Borel Compact.

Your mistake is saying that it is the union of infinitely many intervals. Indeed, the cantor set doesn’t contain any interval!

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  • $\begingroup$ Could you please review my new understanding? Thank you in advance. $\endgroup$
    – ssds
    Sep 4, 2023 at 13:43

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