# Closure of Subset in Subspace

Jump to navigation
Jump to search

## Theorem

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

Let $H$ be a subset of $S$.

Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.

Let $A$ be a subset of $H$.

Then:

- $\map {\cl_H} A = H \cap \map \cl A$

where

## Proof

\(\displaystyle \map {\cl_H} A\) | \(=\) | \(\displaystyle \bigcap \set {K \subseteq H: A \subseteq K, K \text{ is closed in } T_H}\) | Definition of closure of subset | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \bigcap \set {N \cap H: A \subseteq N, N \text{ is closed in } T}\) | Closed Set in Topological Subspace | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle H \cap \bigcap \set {N: A \subseteq N, N \text{ is closed in } T}\) | Intersection Distributes over Intersection of Family of Sets | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle H \cap \map \cl A\) | Definition of closure of subset |

$\blacksquare$