# questions about the Cantor set is Compact

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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.

• 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" ? Sep 2, 2023 at 9:20
• Cantor set is not a union of intervals. It does not contain any (non-degnerate) interval. Sep 2, 2023 at 9:20
• 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.
– Feng
Sep 2, 2023 at 9:23
• Could you please review my new understanding? Thank you in advance
– ssds
Sep 4, 2023 at 13:49
• 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. Sep 4, 2023 at 15:07