Point in Topological Space is Element of its Neighborhood

Theorem

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

Let $x \in S$.

Let $N$ be a neighborhood of $x$ in $T$.

Then $a \in N$.

That is:

$\forall x \in S: \forall N \in \NN_x: x \in N$

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

Proof

Trivially follows by definition of neighborhood of $a$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N2$