Neighborhood Sub-Basis Criterion for Filter Convergence

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Theorem

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

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

Let $p \in X$.


Then $\mathcal F$ converges to $p$ iff $\mathcal F$ contains as a subset a neighborhood sub-basis at $x$.


Proof

Forward implication

If $\mathcal F$ converges to $p$, then it contains every neighborhood of $p$, and the set of neighborhoods of $p$ is trivially a neighborhood sub-basis at $p$.

$\Box$

Reverse Implication

Let $S_p$ be a neighborhood sub-basis at $p$.

Let $S_p \subseteq \mathcal F$.

Let $N$ be a neighborhood of $p$.

Then by the definition of neighborhood sub-basis, there is a finite $T_N \subseteq S_p$ such that:

$\bigcap T_N \subseteq N$.

Since a filter is closed under finite intersections, $\bigcap T_N \in \mathcal F$.

Then $\bigcap T_N \in \mathcal F$ and $\bigcap T_N \subseteq N$, so by the definition of a filter, $N \in \mathcal F$.

Since $\mathcal F$ contains every neighborhood of $p$, it converges to $p$.

$\blacksquare$