---
tag: 09CJ00H3NW
kind: statement
role: proposition
name: Open sets characterised by neighbourhoods
msc: [54A05]
uses: [0CNQXVBHCX, 01ZQQDCERP, 05GK32H72H]
sources:
  - url: https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)
    locator: Neighbourhood of a point
  - url: https://en.wikipedia.org/wiki/Neighbourhood_(mathematics)
    locator: Topology from neighbourhoods
---

## Statement

Let $(X, \tau)$ be a [[topological space|0CNQXVBHCX]] and let $x \in X$.
A set $N \subseteq X$ is a neighbourhood of $x$ when there is an [[open set|01ZQQDCERP]] $U$ with $x \in U \subseteq N$.
Then a set $A \subseteq X$ is open if and only if $A$ is a neighbourhood of each of its points.

## Remarks

Proof of the converse: for each $x \in A$, choose an open $U_x$ with $x \in U_x \subseteq A$. Then $A = \bigcup_{x \in A} U_x$, which is open as a [[union|05GK32H72H]] of open sets.
A neighbourhood need not be open. For example, $[-1, 1]$ is a neighbourhood of $0$ in $\mathbb{R}$.
In a metric space, $N$ is a neighbourhood of $x$ if and only if $N$ contains a ball $B(x, \varepsilon)$ for some $\varepsilon > 0$.

## Explanations

### L1 human

A neighbourhood of $x$ is a set that contains all points near enough to $x$.
A set is open when it is a neighbourhood of each of its points. Then no point of the set lies on its edge.