Problem: Let $(X,\mathcal{T} )$ be a topological space. We say that $U \subset X$ is open if $U\in \mathcal{T} $. Let $A$ be a subset of $X$. Suppose that for each $x\in A$ there is an open set $U \ni x$ such that $U \subset X$. Show that $A$ is open in $X$.
Solution: Given that for any $x\in A$, there exists $U_x \in \mathcal{T} $ such that $U_x \subset A$. We claim that \[ A = \bigcup_{x\in A } U_x \coloneqq \mathcal{U}. \] Let $y\in A$. Then from the given hypothesis $y\in U_y$ and hence $y \in \mathcal{U} $. Therefore, $A \subset \mathcal{U} $. On the other hand if $y \in \mathcal{U} $, then there exists $x \in A$ such that $y \in U_x$. Since $U_x \subset A$, so $y \in A$ and hence $\mathcal{U} \subset A$. Therefore, $A = \mathcal{U} $. Now for each $x\in A$ the set $U_x$ is open and arbitrary union of open set is open, therefore, $A$ is open.