Open sets characterised by neighbourhoods

Statement

Let $(X, \tau)$ be a topological space and let $x \in X$. A set $N \subseteq X$ is a neighbourhood of $x$ when there is an open set $U$ 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 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.