Neighborhood Filter of Point is Filter

Theorem

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

Let $x \in S$ be a point of $T$.

Let $\NN_x$ denote the neighborhood filter of $x$ in $T$.

Then $\NN_x$ is a filter on $S$.

Proof

By definition, $\NN_x$ is the set of all neighborhoods of $x$ in $T$.

It is to be demonstrated that all the conditions are satisfied for $\NN_x$ to be a filter.

Proof of $(\text F 1)$

From Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$.

Hence:

$S \in \NN_x$

Hence $(\text F 1)$ is fulfilled by $\NN_x$.

$\Box$

Proof of $(\text F 2)$

Let $N$ be an element of $\NN_x$

By definition, $x \in N$.

That is:

$N \ne \O$

and it follows that:

$\O \notin \NN_x$

Hence $(\text F 2)$ is fulfilled by $\NN_x$.

$\Box$

Proof of $(\text F 3)$

Let $N_1, N_2 \in \NN_x$.

From Intersection of Neighborhoods in Topological Space is Neighborhood:

$N_1 \cap N_2 \in \NN_x$

Hence $(\text F 3)$ is fulfilled by $\NN_x$.

$\Box$

Proof of $(\text F 4)$

Let $N \in \NN_x$.

Let $M \subseteq S$ such that $N \subseteq M$.

From Superset of Neighborhood in Topological Space is Neighborhood:

$M \in \NN_x$

Hence $(\text F 4)$ is fulfilled by $\NN_x$.

$\Box$

All criteria are fulfilled, and the result follows.

$\blacksquare$