Answer to Question #267264 in Real Analysis for 123

Question

Show that int(A ∩ B) = int(A) ∩ int(B).

Answer 1

$\text{Suppose x $\in$ int(A $\cap$ B), then there exists $\epsilon >0$ such that $N(x,\epsilon)\subset A \implies $ }\\\text{$N(x,\epsilon) \subset A$ and $N(x,\epsilon) \subset B$, then by definition we have that $x \in int(A)$ and $x \in int(B)$}\\ \implies int(A\cap B) \subseteq int(A) \cap int(B)\\ \text{Now, suppose $y\in int(A) \cap int(B),$ then there exist $\epsilon$ > 0 such that $N(y,\epsilon)\in $ }\\ \text{$int(A) \cap int(B)$}\\ \implies N(y,\epsilon) \subset int(A) \text{ and } N(y,\epsilon) \subset int(B)\\ \implies y \in A \text{ and } y \in B\\ \implies y \in A \cap B\\ \text{We have that $y\in int(A)$ and $y\in int(B)$, then there are $\epsilon_1, \epsilon_2 > 0$ such that $N(y,\epsilon_1)$}\\ \text{$\subset A$ and $N(y,\epsilon_2) \subset B$}\\ \text{Let $\epsilon_0 = \min\{\epsilon_1,\epsilon_2\}$, then}\\ N(y,\epsilon_0) \subset A \cap B\\ \implies y \in int(A \cap B)\\ \implies int(A) \cap int(B) \subseteq int(A\cap B)\\ \therefore int(A\cap B) = int(A) \cap int(B)$