Equivalence of Definitions of Compact Topological Subspace

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $T_H = \left({H, \tau_H}\right)$ 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 $\mathcal C$ be a cover of $H$ by open sets of $T$.

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

Since $\mathcal C$ is a cover of $H$ it follows that $\mathcal C' = \left\{{U \cap H: U \in \mathcal C}\right\}$ is also a cover of $H$.

By hypothesis, there exists a finite subcover $\mathcal F'$ of $\mathcal C'$ for $H$.

Thus let $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ where $U_1, U_2, \ldots, U_n \in \mathcal C$.

It follows that $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ 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 $\mathcal C$ be a cover of $H$ by open sets of $H$.

By definition of the subspace topology, each $V \in \mathcal C$ 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 $\mathcal F = \left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal C$ for $H$.

Thus $\mathcal F' = \left\{{U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}\right\}$ 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