sigma field generated by singletons of real line

[rukawakaede]

Suppose set A=\{\{\omega\}:\omega\in\mathbb{R}\}. What can you say more about it? In particular, on the \sigma(A) the smallest sigma field generated by A, i.e. it is closed under complements/countable intersections or unions and the whole space is in the sigma field.

Clearly, if here \mathbb{R} replace to a finite space \Omega, then \sigma(\Omega)=\mathcal{P}(\Omega) since all subset of \Omega can be written as a countable union of singletons of \Omega.

But it is not true for space which is uncountably infinite like \mathbb{R}.

My initial thought is that \sigma(A) does not contain intervals in \mathbb{R}. However i am not sure if I miss anything?

[micromass]

My guess is that \sigma(A)=\{A\subseteq \mathbb{R}~\vert~A~\text{is countable or}~\mathbb{R}\setminus A~\text{is countable}\}

So it wouldn't contain the intervals...

Edit: Wow, LaTeX is acting weird. And I don't think I typed a mistake. anyway, the second line says what I think the sigma-algebra would be like...

[rukawakaede]

My guess is that \sigma(A)=\{A\subseteq \mathbb{R} \vert A is countable or \mathbb{R}\setminus A is countable\}

So it wouldn't contain the intervals...

Edit: Wow, LaTeX is acting weird. And I don't think I typed a mistake. anyway, the second line says what I think the sigma-algebra would be like...

Thanks! I was looking for an argument for this. Now I know the key word is "countable".

[rukawakaede]

My guess is that \sigma(A)=\{A\subseteq \mathbb{R}~\vert~A~\text{is countable or}~\mathbb{R}\setminus A~\text{is countable}\}

So it wouldn't contain the intervals...

Edit: Wow, LaTeX is acting weird. And I don't think I typed a mistake. anyway, the second line says what I think the sigma-algebra would be like...

Also, I found that the TeX rendering system here does not support \text{...} within the tex block. Refer my previous reply for this.

[micromass]

That's wierd, I never noticed \text{...} didn't work. I'm sure I used it before here... Oh well.

[Landau]

It's not so hard to prove that micromass' guess is correct. Certainly \sigma(A) contains all countable subsets (a countable set is the countable union of its singletons) and cocountable subsets (take complements). On the other hand, it is easy to see that his guess is a sigma-algebra containing A.