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Show that G is a sigma-algebra

  1. Aug 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A σ-algebra G on a set X is a family of subsets of X satisfying:

    1) X[itex]\in[/itex]G
    2)A[itex]\in[/itex]G => C(A)[itex]\in[/itex]G
    3)Aj [itex]\subset[/itex] G => [itex]\bigcup[/itex] Aj [itex]\in[/itex] G

    Show that G = {A[itex]\subset[/itex]X : #A≤N or ≠C(A)≤N}

    # stands for the cardinality of the set.
    2. Relevant equations



    3. The attempt at a solution
    Actually I am not so far in the problem solving because I am stuck at showing the first property. We must have that X[itex]\in[/itex]G. But since G is only the set of proper subsets of X, i.e. doesn't contain X by definition, how can 1) hold?
     
  2. jcsd
  3. Aug 28, 2013 #2
    Maybe ##\subset## indicate "subset" instead of the symbol ##\subseteq##...
     
  4. Aug 28, 2013 #3
    wait what? No the definition clearly states to use proper subsets.
     
  5. Aug 28, 2013 #4

    vela

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    Why are you trying to show the first property holds? Isn't it being given to you as part of a definition? In other words, you can assume G satisfies those three properties in trying to write your proof.
     
  6. Aug 28, 2013 #5
    I see I made a mistake. I meant to write: Show that G = {A⊂X : #A≤N or ≠C(A)≤N} is a sigma-algebra on X.
     
  7. Aug 28, 2013 #6

    Office_Shredder

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    The proper subset symbol is often used to denote improper subset so I wouldn't get too caught up in the details of what they are trying to say there.

    However the typical definition of a sigma-algebra does not say that X is in G, it just says that there exists some subset A of X which is contained in G (i.e. G is not empty).
     
  8. Aug 28, 2013 #7
    Yes, you're right, but if ##A\in G\implies X\setminus A\in G## so, by 3) ##A\cup (X\setminus A)= X\in G ##
     
  9. Aug 28, 2013 #8

    Office_Shredder

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    Oops that's embarassing. Then I retract that point and am sticking with "nobody uses the proper subset symbol and means it unless they explicitly state so, so X is contained in G"
     
  10. Aug 29, 2013 #9
    but what if we can't take a proper subset of G? Shouldn't we allow for the case where you have to take all of G (i.e. an improper subset) if we want the argument in #7 to hold?
     
  11. Aug 29, 2013 #10
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