# Compact in Subspace is Compact in Topological Space

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## Theorem

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

Let $K\subseteq S$ be a subset.

Let $\tau_{K}$ be the subspace topology on $K$.

Let $T'=\left(K,\tau_K\right)$ 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 $\{W_i\}_{i \in J}$ be an open cover of $H$ in $T$.

Note that:

- $H=H\cap K \subseteq \bigcup_{i\in J} W_i \iff H\cap K \subseteq \bigcup_{i\in J} W_i \cap K$.

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

Since $H$ is compact in $T'$:

- $H \subseteq \bigcup_{i=1}^{r} \left(W_i \cap K \right)= \left( \bigcup_{i=1}^{r} W_i \right) \cap K$ for some finite subcover $\{W_i \cap K \}_{i =1}^{r}$.

Note that:

- $H\cap K \subseteq \left(\bigcup_{i=1}^{r} W_i \right) \cap K \iff H\cap K \subseteq \bigcup_{i=1}^{r} W_i $.

Therefore:

- $H\subseteq \bigcup_{i\in J} W_i \implies H \subseteq \bigcup_{i=1}^{r} W_i $.

Any open cover $\{W_i\}_{i \in J}$ has a finite subcover $\{W_i \}_{i =1}^{r}$ in $T$.

$\blacksquare$