Show that G is a sigma-algebra

Homework Statement

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

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

Show that G = {A$\subset$X : #A≤N or ≠C(A)≤N}

# stands for the cardinality of the set.

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$\in$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?

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Maybe ##\subset## indicate "subset" instead of the symbol ##\subseteq##...

wait what? No the definition clearly states to use proper subsets.

vela
Staff Emeritus
Homework Helper
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.

Office_Shredder
Staff Emeritus
Gold Member
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).

[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 ##

Office_Shredder
Staff Emeritus