Set is Closed iff Equals Topological Closure/Proof 2
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Theorem
Let $T$ be a topological space.
Let $H \subseteq T$.
Then $H$ is closed in $T$ if and only if:
- $H = \map \cl H$
Proof
Let $H^\complement$ denote the relative complement of $H$ in $T$.
By definition, we have that $H$ is closed in $T$ if and only if $H^\complement$ is open in $T$.
By Set is Open iff Neighborhood of all its Points, this is equivalent to:
- $\forall x \in H^\complement: \exists U \in \tau: x \in U \subseteq H^\complement$
By Empty Intersection iff Subset of Complement, we have that:
- $U \subseteq H^\complement \iff U \cap H = \O$
By Condition for Point being in Closure, it follows that $H^{\complement}$ is open in $T$ if and only if:
- $\forall x \in H^\complement: x \notin \map \cl H$
By the Rule of Transposition, this is equivalent to $\map \cl H \subseteq H$.
From Set is Subset of its Topological Closure, we have that $H \subseteq \map \cl H$.
By definition of set equality, this is equivalent to:
- $H = \map \cl H$
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.15 \ \text{(a)}$