# Neighborhood in Topological Space has Subset Neighborhood

Jump to navigation
Jump to search

## Theorem

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

Let $x \in S$.

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

Then there exists a neighborhood $N'$ of $x$ such that:

- $(1): \quad N' \subseteq N$
- $(2): \quad N'$ is a neighborhood of each of its points.

That is:

- $\forall x \in S: \forall N \in \mathcal N_x: \exists N' \in \mathcal N_x, N' \subseteq N: \forall y \in N': N' \in \mathcal N_y$

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

## Proof

By definition of neighborhood:

- $\exists U \in \tau: x \in U \subseteq N \subseteq S$

where $U$ is an open set of $T$.

By Set is Open iff Neighborhood of all its Points, $N' = U$ fulfils the conditions of the statement.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 3.3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 5$