# Let $X$ be a closed subset of a compact metric space $M$. Then, $X$ is compact.

Theorem : Let $X$ be a closed subset of a compact metric space $M$. Then, $X$ is compact.

Query : Since, $M$ is compact, then there is a finite collection $F$ of open sets which covers $M$. Hence, $F$ should cover every subset of $M$, whether be it open or closed.

Then, why does the problem define this property only for closed subsets of the compact metric?

• The definition of X being compact is not that X has a finite open cover! – Ben FL Oct 2 '14 at 21:06
• You're confused, You want to show that every open cover of $X$ has a finite subcover – CIJ Oct 2 '14 at 21:11
• @ben and @ Carlos, thank you very much for your comments – MathMan Oct 2 '14 at 21:25

I think I see the confusion here: Let $\bigcup O$ be the union of any collection of open sets such that $M \subseteq \bigcup O$. Then there is a finite subcover of $\bigcup O$ that covers $M$. The key word here is any.

Then as you noted, let $K$ be any closed set. Since $K \subseteq M$, $K$ must also be covered by the finite subcover.

Now, if $H$ is an open subset of $M$, it is still a subset of the finite open cover of $M$. However, by the definition of a compact set, there is always some new collection $O'$ which, if we define it cleverly enough, $H$ does not have a finite subcover.

So, even if an open set is a subset of some finite subcover, there is always another cover from which we cannot obtain a finite subcover.

Example: Consider $(0,1)$. Let $O_x = (x/2,1)$. Then $\bigcup_{x\in (0,1)}O_x$ is a cover, but as an exercise you can prove that there is no finite open subcover.

• Oh interesting. I get it . Thank you very much :-) – MathMan Oct 2 '14 at 21:20
• The statement "However, by the definition of a compact set, there is always some new collection O′ which, if we define it cleverly enough, H does not have a finite subcover." cleared everything :-) – MathMan Oct 2 '14 at 21:21

You misunderstand the definition of compactness of $X$. This says that for every open cover of $X$ we can find a finite subcover that also covers $X$. This reduction has to be possible for every open cover; every space can be covered by just one open set, namely itself, so having a finite cover means nothing at all.

So to your problem: You know $M$ is compact, and $X \subset M$ is closed. To show compactness, start by taking any open cover of $X$. To use compactness of $M$, can you augment this cover of $X$ to one of $M$ so that a finite cover of the larger cover also gives you a finite cover for the original? One that uses the closedness?

• Thank you. Does that mean no open subset of a compact metric space can be ever compact? – MathMan Oct 2 '14 at 21:10
• It can happen. But a compact subspace of a Hausdorff space (this includes all metric spaces, e.g.) is always closed. So it goes 2 ways (say for metric): if the large space is compact, every closed subspace is compact too, and if a subspace is compact, it is closed, regardless of the compactness of the whole space. But sets can be both open and closed. As my old teacher said, sets aren't doors: they can be both open and closed at the same time.. – Henno Brandsma Oct 2 '14 at 21:13
• okay .. So, here's my source of confusion exactly : Suppose $M$ is compact,then for every open cover of $M$ we can find a finite subcover that also covers $M$ . Let this finite subcover be $F$. Then it covers $M$. So, why doesn't it cover every subset of $M$? – MathMan Oct 2 '14 at 21:17
• You have to start with a cover of that subset, not one of $M$. $X$ is compact iff every open cover of $X$ has a finite subcover. So you start with that cover, not one of the large space. – Henno Brandsma Oct 2 '14 at 21:19

General version: a closed subspace $X$ of a compact topological space $M$ is compact.

Proof: Suppose we have a covering $$X = \bigcup_i U_i$$ with $U_i$ open in $X$. Then there exist $A_i$ open in $M$ such that $U_i = X \cap A_i \ \forall \ i$ .

Moreover we have $$M = (M \setminus X) \bigcup ( \bigcup_i A_i )$$ M is compact so $$M = (M \setminus X) \bigcup ( \bigcup_{j = 1}^{n} A_{i_j} )$$ for some $i_j$. This implies $$X = \bigcup_{j = 1}^{n} U_{i_j}$$ and so $X$ is compact.

• Thank you for the answer :-) – MathMan Oct 2 '14 at 21:25

Because it's false for open subsets. For instance, $(0,1)$ is not compact.

The proof: If $F\subset X$ is a closed subspace of $X$, given $A_1,\ldots,A_i, \ldots$ open sets of $F$ such that $F = \cup A_i$, then $X = \cup A_i \cup (X\setminus F)$, but each $A_i = F \cap O_i$, where $O_i$ is a open subset of $X$. Then $X = \cup O_i \cup (X\setminus F)$, hence there is a finite sub-colection $\mathcal{F}$ of $\{X\setminus F, O_1, \ldots, O_n, \ldots\}$ such that $X = \bigcup \mathcal{F}$

Define $\mathcal{G} = \{ F \cap Y : Y \in \mathcal{F}\}$, then it's clear that $F = \bigcup \mathcal{G}$ and $\mathcal{G}$ is a finite sub-colection of $\{A_1,\ldots,A_i, \ldots\}$. QED

Can you see what fails if $F$ is open? $X\setminus O$ won't be open and we can't use the compacity o $X$ in our demonstration.

Since $M$ is compact, for any sequence ${x_n} \ \ s.t. x_n\in X$, there exists a convergent subsequence ${x_n}_k$ of ${x_n} \ \ s.t. \ \ {x_n}_k \to x$, for some $x$ in $M$. Clearly, $x$ is a limit point of $X$, and thus by closeness of $X$, $x \in X$. Since any sequence in $X$ has a convergent subsequence in $X$, $X$ is compact as a result.

• Welcome to MSE. I see that you started answering questions, which is very nice. However the question you are answering already has multiple upvoted answer, one of which is also accepted by the OP. Please make shure that whenever you answer a question, your answer adds something to the thread. – gebruiker Dec 21 '15 at 15:51