Limit Points of Countable Complement Space

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Theorem

Let $T = \left({S, \tau}\right)$ be a countable complement space.


Let $H \subseteq S$ be an uncountable subset of $S$.

Then every point of $S$ is a limit point of $H$.


Let $H \subseteq S$ be an countable or finite subset of $S$.

Then $H$ contains all its limit points.


Proof

Let $U \in \tau$ be any open set of $T$.

Let $H \subseteq S$ be an uncountable.

From Uncountable Subset of Countable Complement Space Intersects Open Sets, we have that $U \cap H$ is uncountable if and only if $H$ is uncountable.

Let $x \in S$.

Then every open set $U$ in $T$ such that $x \in U$ also contains an uncountable number of points of $H$ other than $x$.

Thus, by definition, $x$ is a limit point of $H$.


Now let $H$ be a countable subset of $S$.

By definition $\complement_S \left({H}\right)$ is open in $T$ and so $H$ is closed.

So by Closed Set Equals its Closure we have by definition that if $x$ is a limit point of $H$, then $x \in H$.

$\blacksquare$


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