Problem: Let $X$ be a topological space. Let $A$ be a subset of $X$ . Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subseteq A.$ Show that $A$ is open in $X$. (Recall that a set $U$ is open in $X$ if and only if $U$ belongs to the topology of $X$.)
From the given hypothesis $\forall\ x\in A$ find an open set $U_x$ such that $U_x \subseteq A$.
Finally use that arbitrary union of open sets is open to conclude that $A$ is open.
Solution: Let $\mathcal{T} $ be the corresponding topology on $X$. Let $x \in A$. So, for this $x$, we can find $U_x \in \mathcal{T} $ such that $U_x \subseteq A$. We claim that
\[
A = \bigcup_{x\in A} U_x .
\]
Since for any $x \in A$, $U_x \subseteq A$ so the union wil also be a subset of $A$. For the other side, if $y \in A$m, then $y \in U_y \subseteq \cup_{x\in A} U_x$. Thus, $A \subseteq \cup_{x\in A} U_x$ and hence $A = \cup_{x\in A} U_x$. Since $U_x \in \mathcal{T} $ and hence $\cup_{x\in A} U_x \in \mathcal{T} $. Therefore, $A$ is open in $X$.