Show that G is a sigma-algebra

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Homework Statement


A σ-algebra G on a set X is a family of subsets of X satisfying:

1) X\inG
2)A\inG => C(A)\inG
3)Aj \subset G => \bigcup Aj \in G

Show that G = {A\subsetX : #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\inG. But since G is only the set of proper subsets of X, i.e. doesn't contain X by definition, how can 1) hold?
 
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Maybe ##\subset## indicate "subset" instead of the symbol ##\subseteq##...
 
wait what? No the definition clearly states to use proper subsets.
 
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.
 
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.
 
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).
 
Office_Shredder said:
[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 ##
 
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"
 
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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