Closure of Subset in Subspace
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Theorem
Let $T = \struct{S, \tau}$ be a topological space.
Let $H \subseteq S$ be an arbitrary subset of $S$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.
Let $A \subseteq H$ be an arbitrary subset of $H$.
Then:
- $\map {\cl_H} A = H \cap \map \cl A$
where:
- $\map {\cl_H} A$ denotes the closure of $A$ in $T_H$
- $\map \cl A$ denotes the closure of $A$ in $T$.
Corollary 1
Let $K \subseteq S$.
Let $\map {\cl_T} K$ denote the closure of $K$ in $T$.
Let $\map {\cl_H} {K \cap H}$ denote the closure of $K \cap H$ in $T_H$.
Then:
- $\map {\cl_H} {K \cap H} \subseteq \map {\cl_T} K \cap H$
Corollary 2
Let $H$ be closed in $T$.
Then:
- $\map {\cl_H} A = \map \cl A$
Proof
\(\ds \map {\cl_H} A\) | \(=\) | \(\ds \bigcap \set {K \subseteq H: A \subseteq K, K \text{ is closed in } T_H}\) | Definition of Closure (Topology) | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcap \set {N \cap H: A \subseteq N, N \text{ is closed in } T}\) | Closed Set in Topological Subspace | |||||||||||
\(\ds \) | \(=\) | \(\ds H \cap \bigcap \set {N: A \subseteq N, N \text{ is closed in } T}\) | Set Intersection is Self-Distributive over Family of Sets | |||||||||||
\(\ds \) | \(=\) | \(\ds H \cap \map \cl A\) | Definition of Closure (Topology) |
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 28 \ \text {(i)}$
- 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces