closed subsets of a compact set are compact

Theorem 1.

Suppose $X$ is a topological space. If $K$ is a compact subset of $X$, $C$ is a closed set in $X$, and $C\mathrm{\subseteq }K$, then $C$ is a compact set in $X$.

The below proof follows e.g. (http://planetmath.org/Eg) [3]. A proof based on the finite intersection property is given in [4].

Proof.

Let $I$ be an indexing set and $F=\{V_{\alpha }\mid \alpha \in I\}$ be an arbitrary open cover for $C$. Since $X\setminus C$ is open, it follows that $F$ together with $X\setminus C$ is an open cover for $K$. Thus, $K$ can be covered by a finite number of sets, say, $V_1,\mathrm{\dots },V_N$ from $F$ together with possibly $X\setminus C$. Since $C\subset K$, $V_1,\mathrm{\dots },V_N$ cover $C$, and it follows that $C$ is compact.∎

The following proof uses the finite intersection property (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty).

Proof.

Let $I$ be an indexing set and ${\{A_{\alpha }\}}_{\alpha \in I}$ be a collection of $X$-closed sets contained in $C$ such that, for any finite $J\subseteq I$, ${\displaystyle \bigcap _{\alpha \in J}}{A}_{\alpha }$ is not empty. Recall that, for every $\alpha \in I$, $A_{\alpha }\subseteq C\subseteq K$. Thus, for every $\alpha \in I$, $A_{\alpha }=K\cap A_{\alpha }$. Therefore, ${\{A_{\alpha }\}}_{\alpha \in I}$ are $K$-closed subsets of $K$ (see this page (http://planetmath.org/ClosedSetInASubspace)) such that, for any finite $J\subseteq I$, ${\displaystyle \bigcap _{\alpha \in J}}{A}_{\alpha }$ is not empty. As $K$ is compact, ${\displaystyle \bigcap _{\alpha \in I}}{A}_{\alpha }$ is not empty (again, by this result (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty)).This proves the claim.∎