Let $E_1, E_2 \subseteq X$ be closed set in ($X, \tau$) such that $X = E_1\cup E_2$. Consider $Q \subseteq E_2$ open in the subspace topology for $E_2$ and $B\subseteq E_1$ such that $B\cap E_2 \subseteq Q$. Prove that $B \subseteq \text{int}_X(E_1\cup Q)$ where $\text{int}_x()$ is the interior over ($X, \tau$).
I've tried to prove that $B$ is open in $(X,\tau)$, since as $Q$ is open in that subspace then $\exists A\in\tau$ such that $Q=A\cap E_2$ and then $B\cap E_2\subseteq A\cap E_2=Q\implies B\subseteq A$ but I'm not sure how to.
Best Answer
$E_2^{c} \subseteq E_1$ and $E_2^{c}$ is open. It follows that $E_2^{c} \subseteq int_X(E_1) \subseteq int_X(E_1 \cup Q)$. Take any point $b \in B$. if $b \in E_2^{c}$ we are done. Suppose $b \in E_2$. Then $b \in B \cap E_2 \subseteq Q \subseteq A$. I will leave it to you to check that $A \subseteq (E_1\cup Q)$ which shows that $b$ is an interior point of $E_1\cup Q$.