Lemma 5.5.2. Let $X$ be a set and let $\mathcal{B}$ be a collection of subsets. Assume that $X = \bigcup _{B \in \mathcal{B}} B$ and that given $x \in B_1 \cap B_2$ with $B_1, B_2 \in \mathcal{B}$ there is a $B_3 \in \mathcal{B}$ with $x \in B_3 \subset B_1 \cap B_2$. Then there is a unique topology on $X$ such that $\mathcal{B}$ is a basis for this topology.

Proof. Omitted. $\square$

There are also:

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