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Problem is as follow.

Let X be a compact metric space and A be a closed subset of X. Prove that every sequence in A has a convergent (note: convergent in A) subsequence.

It is from my note. My approach is ...

Since $A\subset X$, $A$ is bounded. I don't know $A$ should be compact. If $A$ is compact, $A$ have a convergent subsequence. As a novice, it is hard.


I've added my proof after viewing answer.

Claim : Every closed subset of a compact space X is compact.

Consider any closed subset A of a compact space X.

Let $\mathcal C = \{U_\alpha \mid \alpha \in I\}$ be any open cover of A.

Then $\{U_\alpha \mid \alpha\in I\}\cup\{A^c\}$ is an open cover of X, because $A^c$ is open.

Since X is compact, $$\exists U_{\alpha_1}, ..., U_{\alpha_n} \text{ such that } X \subset \bigcup_{k=1}^n U_{\alpha_k} \cup A^c$$ This implies that $X = \{ \bigcup_{k=1}^n U_{\alpha_k} \} \cup A^c $. (Is this true?)

Since $A\cap A^c=\emptyset$, $$A\subset \bigcup_{k=1}^n \cup_{\alpha_k}\subset \mathcal C.$$ So every open cover of A has a finite subcover of A.

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  • $\begingroup$ subsequence is the right word $\endgroup$ Apr 16 '14 at 10:36
  • $\begingroup$ Thank you :) I've edited my question $\endgroup$
    – jakeoung
    Apr 16 '14 at 10:37
  • $\begingroup$ Since $X$ is a compact metric space, $A$ is indeed compact. Can you prove this? $\endgroup$
    – MPW
    Apr 16 '14 at 10:40
  • $\begingroup$ Now I cannot prove this, but I'm trying. $\endgroup$
    – jakeoung
    Apr 16 '14 at 10:44
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Your approach is correct, although you don't particularly need the fact the space is bounded. To show that $A$ is compact use the following, much more general hint:

HINT: Show that every closed subset of a compact space is compact.

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  • $\begingroup$ Your hint helps me well. I have a question. Can you check (Is this true?) from my question? $\endgroup$
    – jakeoung
    Apr 16 '14 at 11:38
  • $\begingroup$ Yes, your solution is correct. I was glad to help. $\endgroup$
    – Asaf Karagila
    Apr 16 '14 at 11:40
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One way to do it: let $(a_n)_n$ be a sequence from $A$. As $X$ is compact, there is some $x \in X$ and a subsequence $(a_{n_k})_k$ of $(a_n)_n$ such that $a_{n_k} \rightarrow x$ as $k \rightarrow \infty$. As the limit of of a sequence from $A$, $x \in \overline{A}$, but as $A$ is closed, $x \in A$ and so we are done.

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