Closure in Subspace

Theorem

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

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.

Let $U \subseteq T$ and let $U^-$ be the closure of $U$ in $T$.

Then $U^- \cap H$ is the closure of $U \cap H$ in $T_H$.

That is:

$U^- \cap H = \left({U \cap H}\right)^-$

Proof

Let $\mathbb K$ be defined as:

$\mathbb K := \left\{{K \subseteq S: U \subseteq K, K \text { closed in } T}\right\}$
$\displaystyle U^- = \bigcap \mathbb K$

So we have:

 $\displaystyle U^- \cap H$ $=$ $\displaystyle \left({\bigcap \mathbb K}\right) \cap H$ $\displaystyle$ $=$ $\displaystyle \bigcap_{K \mathop \in \mathbb K} \left({K \cap H}\right)$ Intersection is Idempotent

From Closed Set in Topological Subspace it follows that $K \cap H$ is closed in $T_H$.

Thus $\bigcap_{K \mathop \in \mathbb K} \left({K \cap H}\right)$ is the set of all closed sets of $T_H$ which contain $U \cap H$, that is, $\left({U \cap H}\right)^-$.

Hence the result.

$\blacksquare$