Lemma 5.5.3. Let $X$ be a topological space. Let $\mathcal{B}$ be a basis for the topology on $X$. Let $\mathcal{U} : U = \bigcup _ i U_ i$ be an open covering of $U \subset X$. There exists an open covering $U = \bigcup V_ j$ which is a refinement of $\mathcal{U}$ such that each $V_ j$ is an element of the basis $\mathcal{B}$.

Proof. If $x \in U = \bigcup _{i\in I} U_ i$, there is an $i_ x \in I$ such that $x \in U_{i_ x}$. Thus we have a $B_{i_ x} \in \mathcal{B}$ verifying $x \in B_{i_ x} \subset U_{i_ x}$. Set $J = \{ i_ x | x \in U\}$ and for $j = i_ x \in J$ set $V_ j = B_{i_ x}$. This gives the desired open covering of $U$ by $\{ V_ j\} _{j \in J}$. $\square$

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