A subset of a compact set is compact?

Question

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 $T$ — Thomas Andrews, Oct 13 '12 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. — Mathematics, Oct 13 '12 at 16:01

Asaf Karagila · Answer 1 · 2012-10-13T17:24:57.867

50

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 $\cal U$.

edited Oct 13 '12 at 17:24

answered Oct 13 '12 at 17:15

Asaf Karagila

405,794

Is there an example of an open subset of a compact set that isn't compact? – Josie Thompson Sep 05 '21 at 22:35
2

You mean like $(0,1)$ as a subset of $[0,1]$? – Asaf Karagila Sep 05 '21 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. – Josie Thompson Sep 05 '21 at 23:52
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@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. – Asaf Karagila Dec 18 '21 at 13:25
Why is $U\cap S$ open? – Mangostino Apr 12 '22 at 23:48
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@Mangostino: Because that's the definition of relatively open sets? – Asaf Karagila Apr 12 '22 at 23:48
Sorry I deleted my comments. Your first response was sufficient to understand. – Mangostino Apr 13 '22 at 00:00
what if $\mathcal{V}$ is empty, as it generally is on infinite-dimensional spaces? – Mac Menders Jun 05 '25 at 08:59
@MacMenders: How can it be empty? (The point is that we can consider this entire thing happening in $T$ with the subspace topology. I agree that I could have written it better back when I was still doing my M.Sc. and learning how to write mathematics properly, but I do think it's that big of a deal to merit an edit 13 years later.) – Asaf Karagila Jun 05 '25 at 10:52
@AsafKaragila every compact set on an infinite dimensional space has empty interior. So there exist no open U contained in T. – Mac Menders Jun 07 '25 at 13:34
@MacMenders: Read the comment all the way through, please. – Asaf Karagila Jun 07 '25 at 13:50

Ivo Terek · Answer 2 · 2014-07-25T13:21:39.733

34

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 $\{\mathcal{U}_\alpha\}$.

edited Jul 25 '14 at 13:21

answered Jul 25 '14 at 03:34

Ivo Terek

80,301

5

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)? – Madhav Nakar Nov 29 '19 at 23:45
5

@Madhav Yes, precisely. – Ivo Terek Nov 29 '19 at 23:49
Strictly speaking, a set open in $S$ may not be open in $T$. You need to pass through the subspace topology. – RDL Jun 03 '24 at 07:46

score 16 · Answer 3 · edited Aug 22 '14 at 19:06

16

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.

edited Aug 22 '14 at 19:06

answered Oct 13 '12 at 15:57

J.R.

18,312

But i can find a open subcover, which part goes wrong? – Mathematics Oct 13 '12 at 15:59
See Thomas' comment. – J.R. Oct 13 '12 at 16:00
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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 $S$. – Thomas Andrews Oct 13 '12 at 16:01
3

I think you mean "cannot possibly" when you say "can impossibly". – Austin Mohr Oct 13 '12 at 19:59
1

Is that not correct English? I didn't know that, sorry. – J.R. Oct 13 '12 at 22:46
"Then $S$ is relatively compact". But fixed a metric in a space $X$, compactness doesn't depends on the space $X$. Meaning that if $Y \leq X$ is a subspace, $K \subset Y$ is compact w.r.t. $X$ iff it's compact w.r.t. $Y$. – Ivo Terek Jul 25 '14 at 12:47
@YourAdHere I'm pretty sure it's incorrect, yeah – John Fernley Jul 25 '14 at 14:02
@IvoTerek: Relatively compact means just that the closure of the set is compact. – J.R. Aug 27 '14 at 13:10
Oh. I didn't knew. Thanks! – Ivo Terek Aug 27 '14 at 20:27

score 3 · Answer 4 · answered Dec 01 '14 at 15:13

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. $\blacksquare$

Chi-Yeung Law · Answer 5 · 2017-08-20T12:27:54.147

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.

score 0 · Answer 6 · answered Sep 06 '19 at 09:17

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.

A subset of a compact set is compact?

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