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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