Proposition

Suppose $X$ is a topological space, and $\mathcal{B}$ is a basis for its topology. Show that a subset $U \subset X$ is open if and only if it satisfies the following conditions:

Solution

Suppose that $U$ is open. Since $\mathcal{B}$ is a basis, there exist $\{ B_{\alpha} \} \subset \mathcal{B}$ such that $U = \bigcup B_{\alpha}$. Then for each $p \in U$, there must exist a $B_{\alpha}$ such that $p \in B_{\alpha} \subset U$.

On the other hand, suppose that $\forall p \in U, \exists B \in \mathcal{B}, p \in B \subset U$. For each $p \in U$, let $B_p$ denote a basis element such that $p \in B_p \subset U$. In case there are more than one, we will pick one arbitrarily. Then $\bigcup_{p \in U} B_p = U$, so $U$ is open.