Remark

Let $(U_i)_{i \in I}$ be an open covering of a topological space $X$, and for each $i \in I$ let $\mathfrak{B}_i$ be a base of the topology of the subspace $U_i$ of $X$; then it is clear that $\mathfrak{B} = \bigcup_{i \in I} \mathfrak{B}_i$ is a base of the topology of $X$.