# Closed Set in Topological Subspace

## Contents

## Theorem

Let $T$ be a topological space.

Let $T' \subseteq T$ be a subspace of $T$.

Then $V \subseteq T'$ is closed in $T'$ if and only if $V = T' \cap W$ for some $W$ closed in $T$.

### Corollary

Let subspace $T'$ be closed in $T$.

Then $V \subseteq T'$ is closed in $T'$ if and only if $V$ is closed in $T$.

## Proof

### Necessary Condition

Suppose $V \subseteq T'$ is closed in $T'$.

Then $T' \setminus V$ is open in $T'$ by definition.

So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.

Then:

\(\displaystyle V\) | \(=\) | \(\displaystyle T' \setminus \left({T' \setminus V}\right)\) | Relative Complement of Relative Complement | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle T' \setminus \left({T' \cap U}\right)\) | from above | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle T' \setminus U\) | Set Difference with Intersection is Difference | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle T' \cap \left({T \setminus U}\right)\) |

Thus $T \setminus U$ is closed in $T$.

$\Box$

### Sufficient Condition

Conversely, suppose $V = T' \cap W$ where $W$ closed in $T$.

Then $T' \setminus V = T' \setminus \left({T' \cap W}\right) = T' \cap \left({T \setminus W}\right)$ which is open in $T'$.

So $V$ is closed in $T'$.

$\blacksquare$

## Also see

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $3.7$: Definitions: Proposition $3.7.6$