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A mistake from Rudin analysis?

by jessicaw
Tags: analysis, mistake, rudin
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jessicaw
#1
Oct5-10, 11:22 AM
P: 56
Let [tex] B_n=\cup_{i=1}^n A_i [/tex].
[tex] \overline{B_n}[/tex] is the smallest closed subset containing [tex] B_n[/tex].
Note that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is a closed subset containing [tex] B_n[/tex].
Thus,
[tex] \overline{B_n}\supset \cup_{i=1}^n \overline{A_i}[/tex]


Isn't the truth should be that
[tex] \overline{B_n}[/tex] is the smallest?
How come claim that
[tex]\cup_{i=1}^n \overline{A_i}[/tex] is even smaller?
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Landau
#2
Oct5-10, 12:01 PM
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
[tex] \overline{B_n}\subset \cup_{i=1}^n \overline{A_i}.[/tex]
In fact the reversed inclusion
[tex] \overline{B_n}\supset \cup_{i=1}^n \overline{A_i}[/tex]
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
#3
Oct5-10, 01:32 PM
Sci Advisor
P: 905
For example, we could argue as follows:
[tex]A_i\subseteq \overline{A_i}\ \forall i[/tex]

[tex]\Rightarrow \bigcup_i A_i\subseteq \bigcup_i \overline{A_i}[/tex]

[tex]\Rightarrow \overline{\bigcup_i A_i}\subseteq \overline{\bigcup_i \overline{A_i}}[/tex]

A finite union of closed sets is closed, so if I is finite then

[tex]\overline{\bigcup_i \overline{A_i}}=\bigcup_i \overline{A_i}[/tex]

which proves the reversed inclusion

[tex]\overline{\bigcup_i A_i}\subseteq \bigcup_i \overline{A_i}.[/tex]

However, an infinite union of closed sets is not necessarily closed, making this argument stop working. Indeed, consider

[tex]I=\mathbb{Q},\ A_q=\{q\}.[/tex]

Then

[tex]\overline{A_q}=\overline{\{q\}}=\{q\}[/tex].

Hence

[tex]\overline{\bigcup_{q\in I}A_q}=\overline{\mathbb{Q}}=\mathbb{R}[/tex]

[tex]\bigcup_{q\in I}\overline{\{q\}}=\mathbb{Q}.[/tex]


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