Set is Closed iff Equals Topological Closure

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.


Then $H$ is closed in $T$ if and only if:

$H = \operatorname{cl} \left({H}\right)$


That is, a closed set equals its closure (which makes semantic sense).


Proof 1

Let $H'$ denote the derived set of $H$.


By Closed Set iff Contains all its Limit Points, $H$ is closed in $T$ if and only if $H' \subseteq H$.

By Union with Superset is Superset, $H' \subseteq H$ if and only if $H = H \cup H'$.

The result follows from the definition of closure.

$\blacksquare$


Proof 2

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:

$U \subseteq H^{\complement} \iff U \cap H = \varnothing$

By Condition for Point being in Closure, it follows that $H^{\complement}$ is open in $T$ if and only if:

$\forall x \in H^{\complement}: x \notin \operatorname{cl} \left({H}\right)$

By the Rule of Transposition, this is equivalent to $\operatorname{cl} \left({H}\right) \subseteq H$.

From Set is Subset of its Topological Closure, we have that $H \subseteq \operatorname{cl} \left({H}\right)$.

By definition of set equality, this is equivalent to:

$H = \operatorname{cl} \left({H}\right)$

$\blacksquare$