Equivalence of Definitions of Compact Topological Subspace

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$, where $H \subseteq S$.

$1$ implies $2$
Suppose $T_H$ is compact in the sense of Definition 1.

Let $\CC$ be a cover of $H$ by open sets of $T$.

Then for each $U \in \CC$, $U \cap H$ is open in $T_H$ by definition of the subspace topology.

Since $\CC$ is a cover of $H$ it follows that $\CC' = \set{U \cap H: U \in \CC}$ is also a cover of $H$.

By hypothesis, there exists a finite subcover $\FF'$ of $\CC'$ for $H$.

Thus let $\FF' = \set {U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}$ where $U_1, U_2, \ldots, U_n \in \CC$.

It follows that $\FF = \set {U_1, U_2, \ldots, U_n}$ is a finite subcover of $H$.

That is, $T_H$ is compact in the sense of Definition 2.

$2$ implies $1$
Suppose $T_H$ is compact in the sense of Definition 2.

Let $\CC$ be a cover of $H$ by open sets of $H$.

By definition of the subspace topology, each $V \in \CC$ is of the form $U \cap H$ for some $U \in \tau$.

It follows that the collection of all such $U$ is a cover of $H$ by open sets of $T$.

By hypothesis, there exists a finite subcover $\FF = \set {U_1, U_2, \ldots, U_n}$ of $\CC$ for $H$.

Thus $\FF' = \set {U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}$ is also a finite subcover of $H$

That is, $T_H$ is compact in the sense of Definition 1.

Also see

 * Definition:Compact Subspace
 * Definition:Compact Space