# A subset of a compact set is compact?

Claim:Let $S\subset T\subset X$ where $X$ is a metric space. If $T$ is compact in $X$ then $S$ is also compact in $X$.

Proof:Given that $T$ is compact in $X$ then any open cover of T, there is a finite open subcover, denote it as $\left \{V_i \right \}_{i=1}^{N}$. Since $S\subset T\subset \left \{V_i \right \}_{i=1}^{N}$ so $\left \{V_i \right \}_{i=1}^{N}$ also covers $S$ and hence $S$ is compact in X

Edited: I see why this is false but in general, why every closed subset of a compact set is compact?

• You need to prove it for any open cover of $S$, not just the covers of $S$ that also cover $T$ Oct 13, 2012 at 15:58
• O i see which part goes wrong, since $S\subset T$ there may exist a subcover of $S$ which may not cover $T$ and that subcover may not finite at all. Oct 13, 2012 at 16:01
• No, there may exist a coverof $S$ that does not cover $T$ Oct 13, 2012 at 16:05

If $S\subseteq T$ and $T$ is compact and $S$ is closed then $S$ is compact.

Why? Let $\cal U$ be an open cover of $S$. Every open set in $\cal U$ is of the form $U\cap S$ for some open set $U$ (open in $T$). Let $\mathcal V=\{U\subseteq T\mid U\text{ is open, and }\exists U'\in\mathcal U:U\cap S=U'\}$. Then $\mathcal V$ is an open cover of $S$ as well, since $S$ is closed we have that $T\setminus S$ is open so $\mathcal V\cup\{T\setminus S\}$ is an open cover of $T$.

By compactness of $T$ we have a finite subcover, from which we can produce a finite subcover of $\cal U$.

We have shown that every open cover of $S$ has a finite subcover, and therefore $S$ is compact. We have used the fact that $S$ is closed to make sure that $T\setminus S$ is open. If $S$ is not closed we cannot use this to produce an open cover of $T$ and we cannot continue and find an open subcover for $\cal U$.

• Is there an example of an open subset of a compact set that isn't compact? Sep 5, 2021 at 22:35
• You mean like $(0,1)$ as a subset of $[0,1]$? Sep 5, 2021 at 22:46
• yeah, I'm starting to see it now after doing some related exercises, thank you. That probably seemed obvious to you, but as someone who just learned the definition of compact, it wasn't immediately clear to me why (0, 1) isn't compact. In particular, exercises 2.12 and 2.14 from baby Rudin (Principles of Mathematical Analysis by Walter Rudin) are starting to help me understand it. Sep 5, 2021 at 23:52
• @ColmBhandal: It's the definition of subspace topology. The set $(1,4)$ is not an open subset of $[2,3]$, but $(1,4)\cap[2,3]$ is. Dec 18, 2021 at 13:25
• @Mangostino: Because that's the definition of relatively open sets? Apr 12, 2022 at 23:48

Edited: I see why this is false but in general, why every closed subset of a compact set is compact?

Another proof: Let $S \subset T$ be a closed set, where $T$ is compact. Let $\{\mathcal{U}_\alpha\}$ be an open cover of $S$. Then $\{\mathcal{U}_\alpha\} \cup \{S^c\}$, where $S^c$ is the complement of $S$ w.r.t. to $X$, covers $T$. Since $T$ is compact, we can extract a finite subcover $\{ \mathcal{U}_{\alpha_1}, \mathcal{U}_{\alpha_2}, \ldots, \mathcal{U}_{\alpha_n}, S^c \}$ from $\{\mathcal{U}_\alpha\} \cup \{S^c\}$. Notice that $S^c$ maybe wasn't necessary, but we throw it in anyway. Since $S \cap S^c = \varnothing$, we have that $\{ \mathcal{U}_{\alpha_1}, \mathcal{U}_{\alpha_2}, \ldots, \mathcal{U}_{\alpha_n}\}$ is a subcover of $\{\mathcal{U}_\alpha\}$.

• Where do you use the fact that $S$ is closed? Is it used when $S^c$ is put in the cover (since $S^c$ is open)? Nov 29, 2019 at 23:45
• @Madhav Yes, precisely. Nov 29, 2019 at 23:49

Your proof cannot possibly be correct, because the statement is wrong. Note that if $S$ is not closed, then it cannot possibly be compact. Counterexample: $(1/4,1/2)\subset[0,1]\subset\mathbb{R}$.

The correct statement is: If $S\subset T\subset X$, $S$ closed, $T$ compact. Then $S$ is compact.

Alternatively: $S\subset T\subset X$, $T$ compact. Then $S$ is relatively compact.

• But i can find a open subcover, which part goes wrong? Oct 13, 2012 at 15:59
• See Thomas' comment.
– J.R.
Oct 13, 2012 at 16:00
• You've proven only that for some open covers of $S$ there is a finite sub-cover. You need to prove that for all covers of $S$. Oct 13, 2012 at 16:01
• I think you mean "cannot possibly" when you say "can impossibly". Oct 13, 2012 at 19:59
• Is that not correct English? I didn't know that, sorry.
– J.R.
Oct 13, 2012 at 22:46

Suppose $F \subset K \subset X$, F is closed relative to X and K is compact. Let $\{V_{\alpha}\}$ be an open cover of F. Now, F being closed implies $F^c$ is open. Therefore, $F^c \cup \{V_{\alpha}\}$ forms an open cover of set K (As any union of collection of open sets is open). But, K is compact that implies there is a finite sub-cover of $F^c \cup \{V_{\alpha}\}$ denoted by $\beta$ that covers K. Now, $F \subset K$ implies that $\beta$ is a finite cover of F too. Finally, if $F^c \in \beta$ remove it to get a finite sub-cover of $\{V_{\alpha}\}$ that clearly still covers F. Hence, we showed that for any open cover of F denoted by $\{V_{\alpha}\}$ there is a finite sub-cover that covers F. $\blacksquare$

According to the definition of the compact set, we need every open cover of set K contains a finite subcover. Hence, not every subsets of compact sets are compact.

Why closed subsets of compact sets are compact?

Proof

Suppose $F\subset K\subset X$, F is closed in X, and K is compact. Let $\{G_{\alpha}\}$ be an open cover of F. Because F is closed, then $F^{c}$ is open. If we add $F^{c}$ to $\{G_{\alpha}\}$, then we can get a open cover $\Omega$ of K. Because K is compact, $\Omega$ has finite sucollection $\omega$ which covers K, and hence F. If $F^{C}$ is in $\omega$, we can remove it from the subcollection. We have found a finite subcollection of $\{G_{\alpha}\}$ covers F.

Here's an alternate proof (for closed subsets, obviously): any net on $$S$$ is a net on $$T$$ and thus has a convergent subnet with limit is in $$T$$ -- but its limit must also be in $$S$$ because $$S$$ is closed.

It's a little tricky because the notion of closed sets and compact sets are intuitively very similar.