Is X uncountable if there exists a B subset of uncountable with X-B uncountable?

[lreyvega]

Hi everyone! I have the following problem

Homework Statement

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

Homework Equations

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 written 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

 

[Billy Bob]

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.

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).