Elementary Classical Analysis, Marsden, 2nd ed, Chapter 2, Problem 12

Problem

Prove the following properties (for subset of $\mathbb{R}^n$).

(a) $\text{int}(\text{int}(A)) = \text{int}(A)$.

(b) $\text{int}(A \cup B) \supset \text{int}(A) \cup \text{int}(B)$

(c) $\text{int}(A \cap B) = \text{int}(A) \cap \text{int}(B)$.

Answer

(a) For $x \in \text{int}(\text{int}(A))$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset \text{int}(A)$, which means $x \in \text{int}(A)$.

For $x \in \text{int}(A)$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset A$. For $y \in D(x,\epsilon)$, $D(y,\epsilon-|x-y|) \subset D(x,\epsilon) \subset A$, which implies $D(x,\epsilon) \subset \text{int}(A)$. Hence, $x \in \text{int}(\text{int}(A))$.

(b) For $x \in \text{int}(A)$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset A \subset A \cup B$  $\Rightarrow$  $x \in \text{int}(A \cup B)$  $\Rightarrow$  $\text{int}(A) \subset \text{int}(A \cup B)$. Similarly, $\text{int}(B) \subset \text{int}(A \cup B)$.

(c) For $x \in \text{int}(A \cap B)$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset A \cap B \subset A$, which implies $x \in \text{int}(A)$. Similarly, $x \in \text{int}(B)$. Hence, $\text{int}(A \cap B) \subset \text{int}(A) \cap \text{int}(B)$.

For $x \in \text{int}(A) \cap \text{int}(B)$, there exists $\epsilon_A,\epsilon_B >0$ such that $D(x,\epsilon_A) \subset A$ and $D(x,\epsilon_B) \subset B$. For $\epsilon = \min(\epsilon_A,\epsilon_B)$, $D(x,\epsilon) \subset A \cap B$, which implies $x \in \text{int}(A \cap B)$. Hence, $\text{int}(A) \cap \text{int}(B) \subset \text{int}(A \cap B)$.

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