# A mistake from Rudin analysis?

by jessicaw
Tags: analysis, mistake, rudin
 P: 56 Let $$B_n=\cup_{i=1}^n A_i$$. $$\overline{B_n}$$ is the smallest closed subset containing $$B_n$$. Note that $$\cup_{i=1}^n \overline{A_i}$$ is a closed subset containing $$B_n$$. Thus, $$\overline{B_n}\supset \cup_{i=1}^n \overline{A_i}$$ Isn't the truth should be that $$\overline{B_n}$$ is the smallest? How come claim that $$\cup_{i=1}^n \overline{A_i}$$ is even smaller?
 Sci Advisor P: 905 I agree; the fact that the union of closures is a closed subset containing B_n, combined with minimality of cl(B_n), gives $$\overline{B_n}\subset \cup_{i=1}^n \overline{A_i}.$$ In fact the reversed inclusion $$\overline{B_n}\supset \cup_{i=1}^n \overline{A_i}$$ also holds if the union is finite (i.e. the closure operation distributes over finite unions), but not if the union is infinite. But that requires a different argument, so I don't know what Rudin is doing (I don't have his book).
 Sci Advisor P: 905 For example, we could argue as follows: $$A_i\subseteq \overline{A_i}\ \forall i$$ $$\Rightarrow \bigcup_i A_i\subseteq \bigcup_i \overline{A_i}$$ $$\Rightarrow \overline{\bigcup_i A_i}\subseteq \overline{\bigcup_i \overline{A_i}}$$ A finite union of closed sets is closed, so if I is finite then $$\overline{\bigcup_i \overline{A_i}}=\bigcup_i \overline{A_i}$$ which proves the reversed inclusion $$\overline{\bigcup_i A_i}\subseteq \bigcup_i \overline{A_i}.$$ However, an infinite union of closed sets is not necessarily closed, making this argument stop working. Indeed, consider $$I=\mathbb{Q},\ A_q=\{q\}.$$ Then $$\overline{A_q}=\overline{\{q\}}=\{q\}$$. Hence $$\overline{\bigcup_{q\in I}A_q}=\overline{\mathbb{Q}}=\mathbb{R}$$ $$\bigcup_{q\in I}\overline{\{q\}}=\mathbb{Q}.$$