# A mistake from Rudin analysis?

1. ### jessicaw

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?

2. ### Landau

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).

3. ### Landau

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

$$\overline{\bigcup_{q\in I}A_q}=\overline{\mathbb{Q}}=\mathbb{R}$$

$$\bigcup_{q\in I}\overline{\{q\}}=\mathbb{Q}.$$

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