Describe $\sigma$-algebra subsets of $[0;1]$ section generated by.. 
Describe $\sigma$-algebra subsets of $[0;1]$ section generated by following sets system that consists of all single-point sets.

However I do have several ideas, I'm rather new to this topic and not really sure whether I'm right, my ideas are the following:
In this task we need to list all the possible elements of $\sigma$-algebra given, so for any point we may take it and may not, so $\{\{x\}, \{[0;1] \setminus x \} | \forall x \in [0;1] \}$.
Is it right?
Any response/solutions is greatly appreciated.
 A: You want to describe a $\sigma$-algebra generated by
$$\mathcal{E} = \left\{\left\{x\right\}:x\in[0,1]\right\}.$$
Let's call that $\sigma(\mathcal{E})$. By the definition of $\sigma$-algebra, $\sigma(\mathcal{E})$ needs to contain the empty set, be closed under complements, and closed under countable union.
Right now, your collection is not a $\sigma$-algebra because it is not closed under countable unions. For example, the set
$$\left\{\frac1n:n\in\mathbb{N}\right\} = \bigcup_n \left\{\frac1n\right\}$$
should be in $\sigma(\mathcal{E})$ but it is not in your collection you described.
For the actual characterization of $\mathcal{M}$, I think what you would end up with is the so-called "countable co-countable" $\sigma$-algebra. That is, we have
$$\mathcal{M}= \left\{ E\subseteq [0,1]:E\text{ is countable or }[0,1]\setminus E\text{ is countable}\right\}$$
as the $\sigma$-algebra generated by single point sets. You should prove the two following claims as exercises (they are pretty straightforward)
Claim 1: $ \mathcal{M}$ is in fact a $\sigma$-algebra.
Claim 2: $\mathcal{E}\subseteq \mathcal{M}$.
From the above claims, it follows that $\mathcal{M}\supseteq \sigma(\mathcal{E})$ because $\sigma(\mathcal{E})$ is the smallest $\sigma$-algebra containing $\mathcal{E}$. Lastly, proving that (also straightforward) $\mathcal{M}\subseteq \sigma(\mathcal{E})$ will give you $\mathcal{M} = \sigma(\mathcal{E})$.
