Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Baby Rudin Problem 2.7

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data

    I've proved that if [itex] B = \bigcup_{i=1}^{\infty} A_{i} [/itex] then [itex] \overline{B} = \bigcup_{i=1}^{\infty} \overline{A_{i}} [/itex] but it should not be right. So could you find errors on my reasoning?

    2. Relevant equations

    3. The attempt at a solution

    Observe [tex] x \in \overline{B} [/tex]

    iff for every [tex] \epsilon>0 \quad B(x;\epsilon) \cap B \neq \emptyset [/tex]

    iff [tex] B(x;\epsilon) \cap \bigcup_{i=1}^{\infty} A_{i} \neq \emptyset [/tex]

    iff [tex] B(x;\epsilon) \cap A_{i_{0}} \neq \emptyset [/tex] for some [itex] i_{0} \in \mathbb{Z}^{+}[/itex]

    iff [tex] x \in \overline{A_{i_{0}}} [/tex]

    iff [tex] x \in \bigcup_{i=1}^{\infty} \overline{A_{i}} [/tex]
  2. jcsd
  3. Jul 19, 2012 #2


    User Avatar
    Gold Member

    Your fourth iff is wrong.

    it should be if $x \in \bar{A_i_0}$ then $B(x,\epsilon)\cap A_{i_0}$.

    I mean if $B(x,\epsilon) \cap A_{i_0}$ \forall \epsilon >0, x can still be outside of A_{i_0}.

    Take another set such A_j such that x is in it but not in A_i_0 but they intersect eachother, and such that whatever nbhd of x we pick it intersects A_i_0.
  4. Jul 19, 2012 #3
    Oh yeah. I think I know what you mean. i_{0} depends on epsilon. Thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook