## A mistake from Rudin analysis?

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?
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 Recognitions: Science Advisor 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).
 Recognitions: Science Advisor 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 $$\overline{\bigcup_{q\in I}A_q}=\overline{\mathbb{Q}}=\mathbb{R}$$ $$\bigcup_{q\in I}\overline{\{q\}}=\mathbb{Q}.$$