Proposition 3

If $X$ is a topological space, then for a set $\mathfrak{B}$ of open subsets of $X$ to be a base of the topology of $X$ it is necessary and sufficient that for each $x \in X$ the set of $V \in \mathfrak{B}$ such that $x \in V$ is a fundamental system of neighbourhoods of $x$.

It is clear that the condition is necessary. Conversely, if it is satisfied, then given any open set $U$ and any $x \in U$ there is an open set $V_x \in \mathfrak{B}$ such that $x \in V_x \subset U$. The union of the sets $V_x$ for $x \in U$ is therefore equal to $U$. This completes the proof.