Show that G is a sigma-algebra

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  • #1
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Homework Statement


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.

Homework Equations





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?
 

Answers and Replies

  • #2
Maybe ##\subset## indicate "subset" instead of the symbol ##\subseteq##...
 
  • #3
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wait what? No the definition clearly states to use proper subsets.
 
  • #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.
 
  • #5
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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.
 
  • #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).
 
  • #7
[omissis]
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).
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 ##
 
  • #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"
 
  • #9
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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?
 

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