Point in Topological Space has Neighborhood

Theorem

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

Let $x \in S$.

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

That is:

$\forall x \in S: \NN_x \ne \O$

where $\NN_x$ is the neighborhood filter of $x$.

Proof

Let $x \in S$.

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

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 1$