Closure of Topological Closure equals Closure

Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Then:

$\map \cl {\map \cl H} = \map \cl H$

where $\cl$ denotes the closure of $H$.

Proof

It follows directly from Set is Subset of its Topological Closure that:

$\map \cl H \subseteq \map \cl {\map \cl H}$

$\Box$

Let $x \in \map \cl {\map \cl H}$.

Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in \map \cl H$.

If we consider $U$ as an open set containing $y$, it follows that:

$U \cap H \ne \O$

Hence $x \in \map \cl H$.

Hence by definition of subset:

$\map \cl {\map \cl H} \subseteq \map \cl H$

$\Box$

Hence the result by definition of set equality.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.15 \ \text{(c)}$