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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?

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    $\begingroup$ You need to prove it for any open cover of $S$, not just the covers of $S$ that also cover $T$ $\endgroup$ – Thomas Andrews Oct 13 '12 at 15:58
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    $\begingroup$ 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. $\endgroup$ – Mathematics Oct 13 '12 at 16:01
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    $\begingroup$ No, there may exist a coverof $S$ that does not cover $T$ $\endgroup$ – Thomas Andrews Oct 13 '12 at 16:05
  • $\begingroup$ Please do your best to use informative titles. As it is right now, it is impossible to figure out what the question is about from the title and tags. $\endgroup$ – Asaf Karagila Oct 13 '12 at 17:15
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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 $\cal U$.

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

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    $\begingroup$ I guess that "Since $S \cup S^c = \varnothing$" is a typo. Did you want to write intersection $S \cap S^c$ there? $\endgroup$ – Martin Sleziak Jul 25 '14 at 7:19
  • $\begingroup$ @MartinSleziak that's right! Thanks for pointing (: $\endgroup$ – Ivo Terek Jul 25 '14 at 10:13
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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.

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  • $\begingroup$ But i can find a open subcover, which part goes wrong? $\endgroup$ – Mathematics Oct 13 '12 at 15:59
  • $\begingroup$ See Thomas' comment. $\endgroup$ – J.R. Oct 13 '12 at 16:00
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    $\begingroup$ 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 $S$. $\endgroup$ – Thomas Andrews Oct 13 '12 at 16:01
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    $\begingroup$ I think you mean "cannot possibly" when you say "can impossibly". $\endgroup$ – Austin Mohr Oct 13 '12 at 19:59
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    $\begingroup$ Is that not correct English? I didn't know that, sorry. $\endgroup$ – J.R. Oct 13 '12 at 22:46
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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. $\blacksquare$

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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.

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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.

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