I was wondering if my approach was correct :The question :Given interval [0,1] and set $A$ = {all finite and countable subsets of the interval} , what is the $\sigma $-algebra generated by $A$ ?

My try:

First , lets look at the 1-element sets - we have ${\omega}_{1}$ sets like this with their complements. If we look at the 2-element set and so on we can easily see that we always have the same relation -Cardinality of the set = ${\omega}_{0}$ and cardinality of the complement is ${\omega}_{1}$.So my assumption is - The $\sigma $-algebra generated from the countable sets is

$\{X\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}|X|={\omega}_{0}\vee |{X}^{c}|={\omega}_{0}\}$

But I failed to prove it -

I tried to assume there exist some set generated by the $\sigma $-algebra with the property :

$|X|={\omega}_{1}\wedge |X|={\omega}_{1}$

Am I correct and I just need to find a way to prove it or I'm wrong ?Thank you in advance.

My try:

First , lets look at the 1-element sets - we have ${\omega}_{1}$ sets like this with their complements. If we look at the 2-element set and so on we can easily see that we always have the same relation -Cardinality of the set = ${\omega}_{0}$ and cardinality of the complement is ${\omega}_{1}$.So my assumption is - The $\sigma $-algebra generated from the countable sets is

$\{X\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}|X|={\omega}_{0}\vee |{X}^{c}|={\omega}_{0}\}$

But I failed to prove it -

I tried to assume there exist some set generated by the $\sigma $-algebra with the property :

$|X|={\omega}_{1}\wedge |X|={\omega}_{1}$

Am I correct and I just need to find a way to prove it or I'm wrong ?Thank you in advance.