I have the following question:
Let $X$ be a topological space and let $A,B \subseteq X$ such that ${\left( {\bar A} \right)^{\text{o}}} = \emptyset = {B^{\text{o}}}$. Show that ${\left( {A \cup B} \right)^{\text{o}}} = \emptyset$.
Am I allowed to say that the empty set is a subset of every set so $\emptyset \subseteq {\left( {A \cup B} \right)^{\text{o}}}$?
I'm struggling with the reverse inclusion: I have the following $ {\left( {A \cup B} \right)^{\text{o}}} \subseteq A \cup B \subseteq \overline {A \cup B} = \bar A \cup \bar B\ $, but I'm not sure where to go from there?
Best Answer
Let $U\subseteq A\cup B$ be open.
Then $V=U\setminus \bar A$ is open and $V\subseteq (A\cup B)\setminus \bar A\subseteq B$, so $V=\emptyset$. Thus, $U\subseteq \bar A$, whence $U=\emptyset$.
Note that both conditions are necessary: if $A$ is any set such that $\bar A$ has nonempty interior, then $B=\bar A\setminus A$ will have empty interior and $A\cup B=\bar A$ will not.