Set Closure is Smallest Closed Set/Topology
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Theorem
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ in $T$.
Then $H^-$ is the smallest superset of $H$ that is closed in $T$.
Proof
Define:
- $\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.
The claim is that $H^-$ is the smallest set of $\mathbb K$.
From Set is Subset of its Topological Closure:
- $H \subseteq H^-$
From Topological Closure is Closed, $H^-$ is closed in $T$.
Thus $H^- \in \mathbb K$.
Let $K \in \mathbb K$.
From Set Closure as Intersection of Closed Sets:
- $\ds H^- = \bigcap \mathbb K$
Therefore, from Intersection is Subset: General Result:
- $H^- \subseteq K$
Thus by definition $H^-$ is the smallest set of $\mathbb K$.
$\blacksquare$
Sources
- 1953: Walter Rudin: Principles of Mathematical Analysis ... (previous) ... (next): $2.27 c$
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors