Closure of Topological Closure equals Closure

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Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.


Then:

$\left({H^-}\right)^- = H^-$

where $H^-$ denotes the closure of $H$.


Proof

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

$H^- \subseteq \left({H^-}\right)^-$

$\Box$


Let $x \in \left({H^-}\right)^-$.

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

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

$U \cap H \ne \varnothing$

Hence $x \in H^-$.

$\blacksquare$


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