> [!NOTE] **Proposition** > Let $\mathcal{T}$ be a [[Topology|topology]] on a set $X$. A subset $\mathcal{B}$ of $\mathcal{T}$ is a [[Topological Bases|base]] for $\mathcal{T}$ iff the following conditions are met: > (B1) $\mathcal{B}$ is a (open) [[Set covers|cover]] of $X$. > (B2) For any $B_{1}, B_{2} \in \mathcal{B}$ and a point $x\in B_{1} \cap B_{2}$, there is a $B \in \mathcal{B}$ such that $B \subset B_{1} \cap B_{2}$ and $x\in B$. ###### Proof If there is such a topology, then by the definition of a basis, the sets in this topology consist of all possible unions of sets from $\mathcal{B}$. So we only need check that if $\mathcal{T}$ consists of all unions of sets from $\mathcal{B}$ then this is indeed a topology on $X$. We check properties (T1-3). (T1): $X$ is the union of sets from $\mathcal{B}$ by (B1). (T2): If $U, V \in \mathcal{T}$ then $U=\bigcup_{i \in \mathcal{I}} B_i$ and $V=\bigcup_{j \in \mathcal{J}} D_j$, with $B_i, D_j \in \mathcal{B}$. Then $ U \cap V=\bigcup_{(i, j) \in \mathcal{I} \times \mathcal{J}} B_i \cap D_j $ which is a union of sets in $\mathcal{B}$ by (B2), and hence an element of $\mathcal{T}$. This can be extended inductively to any finite intersection. (T3): Any union of unions of sets from $\mathcal{B}$ is a union of sets from $\mathcal{B}$. **Remark**: given a base $\mathcal{B}$, $\mathcal{T}$ is the smallest topology containing $\mathcal{B}$.