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?

Solution 1:

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 $\cal U$.

Solution 2:

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

Solution 3:

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.

Solution 4:

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. $\blacksquare$