Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 7

Problem

Let $A_1,A_2,A_3,\cdots$ be subsets of a metric space.

(a) If $B_n=\bigcup_{i=1}^n A_i$, prove that $\overline{B_n} = \bigcup_{i=1}^n \overline{A_i}$, for $n=1,2,3,\cdots$.

(b) If $B=\bigcup_{i=1}^\infty A_i$, prove that $\overline{B} \supset \bigcup_{i=1}^\infty \overline{A_i}$.

Answer

(a)

($\subset$) Let $x \in \overline{B_n} = B_n \cup (B_n)'$. If $x \in B_n$ then $x \in \bigcup_{i=1}^n A_i \subset \bigcup_{i=1}^n \overline{A_i}$. If $x \in (B_n)'$, to prove by contradiction, suppose $x \notin \overline{A_i}$ for all $i=1,\cdots,n$. Then there exists $\epsilon_i >0$ such that $N_{\epsilon_i}(x) \cap A_i = \emptyset$. For $\epsilon=\min(\epsilon_1,\cdots,\epsilon_n)$, $N_{\epsilon}(x) \cap B_n = \emptyset$, which is contradiction.

($\supset$) Let $x \in \overline{A_i}$ for some $i=1,\cdots,n$. If $x \in A_i$ then $x \in B_n$. If $x \in (A_i)'$ then for $\epsilon >0$, there exists $y \in A_i$ such that $y \in N_\epsilon(x)$ and $y \neq x$. Since $y$ is also contained in $B_n$, $x \in (B_n)'$.

(b) Let $x \in \overline{A_i}$ for some $i=1,2,\cdots$. If $x \in A_i$ then $x \in B_n$. If $x \in (A_i)'$ then for $\epsilon >0$, there exists $y \in A_i$ such that $y \in N_\epsilon(x)$ and $y \neq x$. Since $y$ is also contained in $B_n$, $x \in (B_n)'$.

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