A *base* (or *basis*) for a [[Topology|topology]] $\mathcal{T}$ on a set $X$ is a subset $\mathcal{B} \subset \mathcal{T}$ such that every set in $\mathcal{T}$ is the union of some sets from $\mathcal{B}$, i.e. for all $U \in \mathcal{T}$, there exists $\mathcal{C}_U \subset \mathcal{B}$ such that $U=\bigcup_{B \in C_U} B$. **Example**: Singleton sets form a base for the discrete topology. **Example**: $\{ (n, n+2) \mid n\in \mathbb{Z} \}$ is an open cover for $\mathbb{R}$ but is clearly not a basis for any topology on $\mathbb{R}$. # Statement > [!NOTE] **Proposition** > A subset $\mathcal{B}$ of subsets of a set $X$ is a basis for some topology of $X$ 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 $x\in B$. **Remark**: If $\mathcal{B} \subset \mathcal{T}$ then $\mathcal{T}$ the smallest topology containing $\mathcal{B}$. # Proof (Moreira, MA260 Norms, metrics and topologies) We check that if $\mathcal{T}$ consists of all unions of sets from $\mathcal{B}$ then this is indeed a topology on $X$. (T1): The empty union gives $\emptyset$ and by (B1), $X\in \mathcal{T}$. (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 $\mathcal{B}$.