Compact in Subspace is Compact in Topological Space

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Theorem

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

Let $K \subseteq S$ be a subset.

Let $\tau_K$ be the subspace topology on $K$.

Let $T' = \struct {K, \tau_K}$ be the topological subspace of $T$ determined by $K$.

Let $H \subseteq K$ be compact in $T'$.


Then $H$ is compact in $T$.


Proof

Suppose that $H$ is compact in $T'$.

Let $\set {W_i}_{i \mathop \in J}$ be an open cover of $H$ in $T$.

Then $\ds H \subseteq \bigcup_{i \mathop \in J} W_i$.

Then:

\(\ds H\) \(=\) \(\ds H \cap K\) Intersection with Subset is Subset: from $H \subseteq K$
\(\ds \) \(\subseteq\) \(\ds \paren {\bigcup_{i \mathop \in J} W_i} \cap K\) Set Intersection Preserves Subsets: Corollary: from $\ds H \subseteq \bigcup_{i \mathop \in J} W_i$
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop \in J} \paren {W_i \cap K}\) Intersection Distributes over Union of Family

Since $W_i \cap K \in \tau_K$, $\set {W_i \cap K}_{i \mathop \in J}$ is an open cover of $H$ in $T'$.


Since $H$ is compact in $T'$, $\set {W_i \cap K}_{i \mathop \in J}$ has some finite subcover $\set {W_i \cap K}_{i \mathop = 1}^r$.

Therefore:

\(\ds H\) \(\subseteq\) \(\ds \set {W_i \cap K}_{i \mathop = 1}^r\) Definition of Finite Subcover
\(\ds \) \(\subseteq\) \(\ds \paren {\bigcup_{i \mathop = 1}^r W_i} \cap K\) Intersection Distributes over Union of Family
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop = 1}^r W_i\) Intersection is Subset


So $\set {W_i}_{i \mathop = 1}^r$ is an open cover of $H$ in $T$, which is a finite subcover.

As $\set {W_i}_{i \mathop \in J}$ is arbitrary:

Any open cover of $H$ has a finite subcover in $T$.

So $H$ is compact in $T$.

$\blacksquare$