# Closure in Subspace

## Theorem

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

Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.

Let $U \subseteq H$ and let $\map {\cl_T} U$ be the closure of $U$ in $T$.

Then $\map {\cl_T} U \cap H$ is the closure of $U$ in $T_H$.

That is:

$\map {\cl_T} U \cap H = \map {\cl_H} U$

where $\map {\cl_H} U$ denotes the closure of $U$ in $T_H$.

### Corollary

Let $W \subseteq S$ and let $\map {\cl_T} W$ denote the closure of $W$ in $T$.

Let $\map {\cl_H} {W \cap H}$ denote the closure of $W \cap H$ in $T_H$.

Then:

$\map {\cl_H} {W \cap H} \subseteq \map {\cl_T} W \cap H$

## Proof

Let $\mathbb K$ be defined as:

$\mathbb K := \set{K \subseteq S: U \subseteq K, K \text { closed in } T}$
$\displaystyle \map {\cl_T} U = \bigcap \mathbb K$

Then:

 $\displaystyle \map {\cl_T} U \cap H$ $=$ $\displaystyle \paren{\bigcap \mathbb K} \cap H$ $\displaystyle$ $=$ $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren{K \cap H}$ Intersection is Idempotent

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

Let $K'$ be a closed set in $T_H$ containing $U$.

From Closed Set in Topological Subspace, there exists a closed set $K$ in $T$ such that $K' = K \cap H$.

Since $U \subseteq K' \subseteq K$ then $K \in \mathbb K$.

Thus $\set{K \cap H : K \in \mathbb K}$ is the set of all closed sets of $T_H$ which contain $U$.

From Set Closure as Intersection of Closed Sets $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren {K \cap H}$ is the closure of $U$ in $T_H$.

The result follows.

$\blacksquare$