# Show that G is a sigma-algebra

1. Aug 28, 2013

### aaaa202

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$\in$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.
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$\in$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?

2. Aug 28, 2013

### Mathitalian

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

3. Aug 28, 2013

### aaaa202

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

4. Aug 28, 2013

### vela

Staff Emeritus
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. Aug 28, 2013

### aaaa202

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. Aug 28, 2013

### Office_Shredder

Staff Emeritus
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. Aug 28, 2013

### Mathitalian

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. Aug 28, 2013

### Office_Shredder

Staff Emeritus
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. Aug 29, 2013

### aaaa202

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?

10. Aug 29, 2013