Theorem.
Let $X$ be a topological space and $Y$ a subset of $X$ . Then the following statements are equivalent.

1. $Y$ is compact as a subset of $X$ .
2. Every open cover of $Y$ (with open sets in $X$ ) has a finite subcover.

Proof. Suppose $Y$ is compact, and $\{U_i\}_{i\in I}$ is an arbitrary open cover of $Y$ , where $U_i$ are open sets in $X$ . Then $\{U_i\cap Y\}_{i\in I}$ is a collection of open sets in $Y$ with union $Y$ . Since $Y$ is compact, there is a finite subset $J\subset I$ such that $Y=\cup_{i\in J} (U_i\cap Y)$ . Now $Y=(\cup_{i\in J} U_i)\cap Y \subset \cup_{i\in J} U_i$ , so $\{U_i\}_{i\in J}$ is finite open cover of $Y$ .

Conversely, suppose every open cover of $Y$ has a finite subcover, and $\{U_i\}_{i\in I}$ is an arbitrary collection of open sets (in $Y$ ) with union $Y$ . By the definition of the subspace topology, each $U_i$ is of the form $U_i = V_i\cap Y$ for some open set $V_i$ in $X$ . Now $U_i \subset V_i$ , so $\{V_i\}_{i\in I}$ is a cover of $Y$ by open sets in $X$ . By assumption, it has a finite subcover $\{V_i\}_{i\in J}$ . It follows that $\{U_i\}_{i\in J}$ covers $Y$ , and $Y$ is compact.

The above proof follows the proof given in [1].

Bibliography

1

B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/~bikenaga/topology/topnote.html.