I have the following question:
Let $X$ be a topological space and $A,B \subseteq X$ such that ${\left( {\bar A} \right)^{\text{o}}} = A{\text{ }}$ and ${\left( {\bar B} \right)^{\text{o}}} = B$.
Prove that:(i) ${\left( {\overline {A \cap B} } \right)^{\text{o}}} = A \cap B$
(ii) $A \subseteq B \Leftrightarrow \bar A \subseteq \bar B$
$Note:{\text{ Here (}} \cdot {{\text{)}}^{\text{o}}}{\text{ is the interior operator and }}\overline {( \cdot )} {\text{ is the closure operator}}{\text{.}}$
My work so far:
(i)
$\begin{gathered}
A \cap B = {\left( {\bar A} \right)^{\text{o}}} \cap {\left( {\bar B} \right)^{\text{o}}} = {\left( {\bar A \cap \bar B} \right)^{\text{o}}} \supseteq {\left( {\overline {A \cap B} } \right)^{\text{o}}} \hfill \\
\therefore {\left( {\overline {A \cap B} } \right)^{\text{o}}} \subseteq A \cap B \hfill \\
\end{gathered} $
How can I show the reverse inclusion?
(ii) $A \subseteq B \Rightarrow \bar A \subseteq \bar B$ is easy to establish, I'm struggling mostly with the converse ($\bar A \subseteq \bar B \Rightarrow A \subseteq B$).
If I start with $\bar A \subseteq \bar B \Rightarrow \overline {{{\left( {\bar A} \right)}^{\text{o}}}} \subseteq \overline {{{\left( {\bar B} \right)}^{\text{o}}}}$, how should this reducte to $A \subseteq B$?
Best Answer
(i)
$A$ and $B$ are interiors of sets, hence are open sets.
Then also $A\cap B$ is an open set, and this with $A\cap B\subseteq\overline{A\cap B}$.
This implies that $A\cap B\subseteq\left(\overline{A\cap B}\right)^{\circ}$
(ii)
From $\overline{A}\subseteq\overline{B}$ it follows directly that $\left(\overline{A}\right)^{\circ}\subseteq\left(\overline{B}\right)^{\circ}$ which in this context is the same as $A\subseteq B$.