Is There a Mistake in Rudin's Analysis Regarding Closed Sets?

[jessicaw]

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?

 

[Landau]

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).

 

[Landau]

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}.