Closure of Subset in Subspace/Corollary 2
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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 $3.9: 28 \ \text {(ii)}$