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Sigma-field on countable set

  1. Apr 14, 2009 #1
    Hi everyone!! I have the following problem

    1. The problem statement, all variables and given/known data

    X is countable set if and only if the sigma-field 2[tex]_{X}[/tex] is generated by the sets of the form [tex]\left\{[/tex][tex]\omega[/tex][tex]\right\}[/tex] with [tex]\omega[/tex][tex]\in[/tex] X


    2. Relevant equations



    3. The attempt at a solution

    The easy part is X is countable then...
    I'm trying to prove the other part. So, I know that the sigma-field 2[tex]_{X}[/tex] is generated by the sets of the form [tex]\left\{[/tex][tex]\omega[/tex][tex]\right\}[/tex] with [tex]\omega[/tex][tex]\in[/tex] X. This means that 2[tex]_{X}[/tex] can be writen as

    2[tex]_{X}[/tex]=[tex]\cap[/tex] [tex]\sigma[/tex][tex]_{\alpha}[/tex]

    where [tex]\sigma[/tex][tex]_{\alpha}[/tex] are all the sigma-fields that contain the family of the singletons. But 2[tex]_{X}[/tex] is the biggest sigma-field that can be constructed in X. That means that [tex]\sigma[/tex][tex]_{\alpha}[/tex]=2[tex]_{X}[/tex] [tex]\forall[/tex]
    [tex]\alpha[/tex].

    Now suppose that X is uncountable. Be the sigma-field [tex]\sigma[/tex] constructed in the following manner:

    A[tex]\subseteq[/tex] X is in the [tex]\sigma[/tex] if A is countable or finite or if X\A is countable or finite.

    It is clear that this sigma-field contains the class formed by the singletons in X. However, this class it is not 2[tex]_{X}[/tex], because if X is uncountable exists B[tex]\subset[/tex] uncountable with X\B uncountable. That is, B is not in [tex]\sigma[/tex]. This means that

    2[tex]_{X}[/tex] is not equal to [tex]\cap[/tex] [tex]\sigma[/tex][tex]_{\alpha}[/tex]

    So, this is a contradiction and X is numerable.

    So, my questions are:

    1) Is this proof correct?
    2) If it is correct, I think that the critical point in it is this fact I used:

    If X is uncountable exists B[tex]\subset[/tex] uncountable with X\B uncountable

    How can I prove this? I think that I need to use the axiom of choice, but I have no clue how to do it.

    Thanks and sorry for my bad english
     
  2. jcsd
  3. Apr 16, 2009 #2
    The rest of your proof looks good. To prove this step, well-order X (OK by axiom of choice).

    Now let B be the set of points b in X that are not of the form b=a+1, for a in X.

    Then B is uncountable (assume not and deduce a contradiction).

    Then X-B is clearly uncountable (for each b in B, we have b+1 in X-B).
     
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