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.**
Let $\sigma (\mathcal B)$ be the set of subsets of $X$ which can be written as unions of elements of $\mathcal B$. We claim $\sigma (\mathcal{B})$ is a topology. Namely, the empty set is an element of $\sigma (\mathcal{B})$ (as an empty union) and $X$ is an element of $\sigma (\mathcal{B})$ (as the union of all elements of $\mathcal{B}$). It is clear that $\sigma (\mathcal{B})$ is preserved under unions. Finally, if $U, V \in \sigma (\mathcal{B})$ then write $U = \bigcup _{i \in I} U_ i$ and $V = \bigcup _{j \in J} V_ j$ with $U_ i, V_ j \in \mathcal{B}$ for all $i \in I$ and $j \in J$. Then

The assumption in the lemma tells us that $U_ i \cap V_ j \in \sigma (\mathcal{B})$ hence we see that $U \cap V$ is too. Thus $\sigma (\mathcal{B}$ is a topology. Properties (1) and (2) of Definition 5.5.1 are immediate for this topology. To prove the uniqueness of this topology let $\mathcal T$ be a topology on $X$ such that $\mathcal B$ is a base for it. Then of course every element of $\mathcal{B}$ is in $\mathcal{T}$ by (1) of Definition 5.5.1 and hence $\sigma (\mathcal{B} \subset \mathcal{T}$. Conversely, part (2) of Definition 5.5.1 tells us that every element of $\mathcal{T}$ is a union of elements of $\mathcal{B}$, i.e., $\mathcal{T} \subset \sigma (\mathcal{B})$. This finishes the proof. $\square$

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