Point in Topological Space has Neighborhood

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

Let $x \in S$.

Then there exists in $T$ at least one neighborhood of $x$.

That is:
 * $\forall x \in S: \mathcal N_x \ne \varnothing$

where $\mathcal N_x$ is the neighborhood filter of $x$.

Proof
Let $x \in S$.

Then $S$ itself is a neighborhood of $x$.