# Sigma-field on countable set

1. Apr 14, 2009

### lreyvega

Hi everyone!! I have the following problem

1. The problem statement, all variables and given/known data

X is countable set if and only if the sigma-field 2$$_{X}$$ is generated by the sets of the form $$\left\{$$$$\omega$$$$\right\}$$ with $$\omega$$$$\in$$ X

2. Relevant equations

3. The attempt at a solution

The easy part is X is countable then...
I'm trying to prove the other part. So, I know that the sigma-field 2$$_{X}$$ is generated by the sets of the form $$\left\{$$$$\omega$$$$\right\}$$ with $$\omega$$$$\in$$ X. This means that 2$$_{X}$$ can be writen as

2$$_{X}$$=$$\cap$$ $$\sigma$$$$_{\alpha}$$

where $$\sigma$$$$_{\alpha}$$ are all the sigma-fields that contain the family of the singletons. But 2$$_{X}$$ is the biggest sigma-field that can be constructed in X. That means that $$\sigma$$$$_{\alpha}$$=2$$_{X}$$ $$\forall$$
$$\alpha$$.

Now suppose that X is uncountable. Be the sigma-field $$\sigma$$ constructed in the following manner:

A$$\subseteq$$ X is in the $$\sigma$$ if A is countable or finite or if X\A is countable or finite.

It is clear that this sigma-field contains the class formed by the singletons in X. However, this class it is not 2$$_{X}$$, because if X is uncountable exists B$$\subset$$ uncountable with X\B uncountable. That is, B is not in $$\sigma$$. This means that

2$$_{X}$$ is not equal to $$\cap$$ $$\sigma$$$$_{\alpha}$$

So, this is a contradiction and X is numerable.

So, my questions are:

1) Is this proof correct?
2) If it is correct, I think that the critical point in it is this fact I used:

If X is uncountable exists B$$\subset$$ uncountable with X\B uncountable

How can I prove this? I think that I need to use the axiom of choice, but I have no clue how to do it.

Thanks and sorry for my bad english

2. Apr 16, 2009

### Billy Bob

The rest of your proof looks good. To prove this step, well-order X (OK by axiom of choice).

Now let B be the set of points b in X that are not of the form b=a+1, for a in X.

Then B is uncountable (assume not and deduce a contradiction).

Then X-B is clearly uncountable (for each b in B, we have b+1 in X-B).