Closed Set in Topological Subspace/Corollary

Theorem

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

Let $T' = \struct{H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.

Let $H$ be closed in $T$.

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

Proof

Let $V \subseteq H$ be closed in $T'$.

Then, from Closed Set in Topological Subspace, $V = H \cap V$ is closed in $T'$.

If $V$ is closed in $T'$ then $V = H \cap W$ where $W$ is closed in $T$.

Since $H$ is closed in $T$, it follows by Topology Defined by Closed Sets that $V$ is closed in $T$.

$\blacksquare$

Also see

• Open Set in Open Subspace

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Corollary $3.7.7$
• 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces