Limit Point iff Superfilter Converges

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\mathcal F$ be a filter on $S$.

Let $x \in S$.


Then $x$ is a limit point of $\mathcal F$ if and only if there exists a superfilter $\mathcal F'$ of $\mathcal F$ on $S$ which converges to $x$.


Proof

Let $x$ be a limit point of $\mathcal F$.

Define:

$\mathcal B := \left\{{F \cap U : F \in \mathcal F \text{ and } U \text{ is a neighborhood of } x}\right\}$

Then $\mathcal B$ is filter basis by definition.

Let $\mathcal F'$ be the corresponding generated filter.

By construction we have $\mathcal F \subseteq \mathcal F'$ and $U \in \mathcal F'$ for every neighborhood $U$ of $x$.

Thus $\mathcal F'$ converges to $x$.

$\Box$


Let there be a filter $\mathcal F'$ on $X$ satisfying $\mathcal F \subseteq \mathcal F'$ which converges to $x$.

Let $U \subseteq X$ be a neighborhood of $x$.

Let $F \in \mathcal F$.

Then:

$U, F \in \mathcal F'$

and therefore:

$U \cap F \in \mathcal F'$

Because $\varnothing \notin \mathcal F'$ it follows that:

$U \cap F \ne \varnothing$

This holds for any neighborhood $U$ of $x$.

Hence $x$ is a limit point of $F$.

Therefore:

$x \in \overline F$

Because this holds for all $F \in \mathcal F$:

$\displaystyle x \in \bigcap \left\{{\overline F : F \in \mathcal F}\right\}$

and so $x$ is a limit point of $\mathcal F$.

$\blacksquare$