prove that $\operatorname{Int}(A \cap B)= \operatorname{Int}(A) \cap \operatorname{Int}(B)$

Question

In the book that I am using they left the proof to the theorem "as a exercise" so I would like to give it a go. Please let me know if I am incorrect.

Let $A$ and $B$ be subsets of a metric space $X$.

$\Rightarrow$

$\operatorname{Int}(A \cap B) \subseteq A \cap B$.

Now as $A \cap B\subseteq A $ and $A \cap B\subseteq B$,

we have $\operatorname{Int}(A \cap B) \subseteq A$ and $\operatorname{Int}(A \cap B) \subseteq B$.

As $\operatorname{Int}(A \cap B)$ is an open set, thus $\operatorname{Int}(A \cap B)\subseteq\operatorname{Int}(A)$ and $\operatorname{Int}(A \cap B)\subseteq\operatorname{Int}(B)$,

thus $\operatorname{Int}(A \cap B)\subseteq\operatorname{Int}(A)\cap \operatorname{Int}(B)$

$\Leftarrow$

As $\operatorname{Int}(A)\subseteq A$ and $\operatorname{Int}(B)\subseteq B$,

thus $\operatorname{Int}(A)\cap\operatorname{Int}(B) \subseteq A\cap B$.

As both $\operatorname{Int}(A)$ and $\operatorname{Int}(B)$ are open sets, and a intersection of finitely many open set is open,

thus $\operatorname{Int}(A)\cap\operatorname{Int}(B)$ itself is open.

Thus $\operatorname{Int}(A)\cap\operatorname{Int}(B) \subseteq\operatorname{Int}(A\cap B)$.

Thus we arrive at the conclusion that $\operatorname{Int}(A \cap B)= \operatorname{Int}(A) \cap\operatorname{Int}(B)$.

Answer 1

Your proof is correct. The first part is also the general form for any interior operator on a power set. You could even leave out a step by directly using monotony to conclude $\operatorname{int}(A\cap B)\subseteq\operatorname{int}(A)$ and $\operatorname{int}(A\cap B)\subseteq\operatorname{int}(B)$ from $A\cap B\subseteq A$ and $A\cap B\subseteq B$.

Using the definition of the interior of topology as the union of all open subsets there is also a direct equation between the two expressions. Let $(X,\mathcal{T})$ be a topological space and $A,B\subseteq X$ be subsets, then: \begin{align*} \operatorname{int}(A)\cap\operatorname{int}(B) &=\left(\bigcup_{\substack{U\subseteq A, \\ U\in\mathcal{T}}}U\right) \cap\left(\bigcup_{\substack{V\subseteq B, \\ V\in\mathcal{T}}}V\right) =\bigcup_{\substack{U\subseteq A, \\ U\in\mathcal{T}}}\left(\bigcup_{\substack{V\subseteq B, \\ V\in\mathcal{T}}}U\right)\cap V =\bigcup_{\substack{U\subseteq A, \\ U\in\mathcal{T}}}\bigcup_{\substack{V\subseteq B, \\ V\in\mathcal{T}}}U\cap V \\ &=\bigcup_{\substack{W\subseteq A\cap B, \\ W\in\mathcal{T}}}W =\operatorname{int}(A\cap B). \end{align*} You also find an answer here.