Compact in Subspace is Compact in Topological Space
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 $T$.
$\blacksquare$