# I was wondering if my approach was correct : The question : Given interval [0,1] and set A = {al

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
$\left\{X\phantom{\rule{1em}{0ex}}|\phantom{\rule{1em}{0ex}}|X|={\omega }_{0}\vee |{X}^{c}|={\omega }_{0}\right\}$
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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Raven Higgins
Not quite: The problem is that your set does not contain complements which are finite.For instance it does not contain the empty set, yet we have: $\left[0,1{\right]}^{c}=\varnothing$.So your set is not a sigma-algebra(since sigma-algebras are closed under complements).
So the answer to your question needs to be slightly tweaked, namely the sigma algebra generated by the set of all countable subsets of [0,1], is the subset $\sigma$ of [0,1] such that elements of $\sigma$ are exactly those elements of [0,1] which are at most countable or are complements of at most countable sets. It’s easy to check that $\sigma$ is a sigma-algebra and that it is the smallest sigma-algebra containing the set of all countable subsets of [0,1].