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A mistake from Rudin analysis? 
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#1
Oct510, 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? 


#2
Oct510, 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). 


#3
Oct510, 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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