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A mistake from Rudin analysis?

  1. Oct 5, 2010 #1
    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].
    [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. jcsd
  3. Oct 5, 2010 #2


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    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).
  4. Oct 5, 2010 #3


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    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]




    [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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