Closure of Subset in Subspace/Corollary 2

Corollary to Closure of Subset in Subspace

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 $H$.

Let $H$ be closed in $T$.

Then:

$\map {\cl_H} A = \map \cl A$

where:

$\map {\cl_H} A$ denotes the closure of $A$ in $T_H$

$\map \cl A$ denotes the closure of $A$ in $T$.

Proof

From Closure of Subset in Subspace:

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

From Intersection is Subset:

$\map {\cl_H} A \subseteq \map \cl A$

From Topological Closure is Closed:

$\map \cl A$ is closed in $T$

From Intersection of Closed Sets is Closed in Topological Space:

$\map {\cl_H} A$ is closed in $T$

From Set is Subset of its Topological Closure:

$A \subseteq \map {\cl_H} A$

From Set Closure is Smallest Closed Set in Topological Space:

$\map \cl A \subseteq \map {\cl_H} A$

By definition of set eqality:

$\map {\cl_H} A = \map \cl A$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 28 \ \text {(ii)}$