Condition for Point being in Closure/Proof 1

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Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Let $H^-$ denote the closure of $H$ in $T$.


Let $x \in S$.

Then $x \in H^-$ if and only if every open neighborhood of $x$ contains a point in $H$.


Proof

From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$.

By definition, an open neighborhood of $x$ in $T$ is an open set of $T$ which contains $x$.


Necessary Condition

Let $x \in H^-$.

Then either:

$(1): \quad x \in H$, in which case every open neighborhood of $x$ in $T$ trivially contains a point in $H$ (that is, $x$ itself);
$(2): \quad x$ is a limit point of $H$ in $T$.


Suppose $(2)$ holds.

Then it follows directly from the definition of limit point that every open neighborhood of $x$ in $T$ contains a point in $H$ other than $x$.


Sufficient Condition

Suppose that every open neighborhood of $x$ in $T$ contains a point in $H$.


If $x \in H$, then $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:

$x \in H^-$


If $x \notin H$ then $x$ must be a limit point of $H$ by definition.

So again, $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:

$x \in H^-$

$\blacksquare$