Set is Open iff Disjoint from Boundary

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Theorem

Let $T$ be a topological space, and let $H \subseteq T$.


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

$\partial H \cap H = \varnothing$

where $\partial H$ is the boundary of $H$.


Proof

From Boundary is Intersection of Closure with Closure of Complement:

$\partial H = H^- \cap \left({T \setminus H}\right)^-$

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

Hence from Intersection is Subset we have that:

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

But from Closed Set Equals its Closure, $\left({T \setminus H}\right)^- = T \setminus H$ if and only if $T \setminus H$ is closed in $T$.

That is, iff $H$ is open in $T$.

So $\partial H \subseteq T \setminus H$ if and only if $H$ is open in $T$.

From Intersection with Complement is Empty iff Subset it follows that $\partial H \cap H = \varnothing$ if and only if $H$ is open in $T$.

$\blacksquare$


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