Fields of sets and sigma-algebras of natural numbers I have the following sets
$$ S=\{\{n\}:n \in \mathbb{N} \}$$
$$ F=\{A \subseteq \mathbb{N}:|A|=2  \}$$
$$ C=\{A \subseteq \mathbb{N}:|A| \leq \aleph_0  \}$$
I denoted by $ c(\cdot)$ field of set generated by this set and by $\sigma(\cdot)$ sigma-algebra generated by this set.
Now, I think that $c(S)=c(F)$ and $c(S)=c(C)$, although I have concerns about the last one - on the other hand $S\subset c(C)$ and so $c(S)\subset c(C)$. Also if $X \in C, |X|=\aleph_0 $ then $|X^{C}|=0$ so it is in $c(S)$ and hence $X\in c(S)$ so $C \subset c(S)$ and $c(C)=c(S)$.
This means that $\sigma(S)=\sigma(F)=\sigma(C)$. I'm not really sure about this. Something tells me that $\sigma(C)=c(C)=P(\mathbb{N})$ (power set) but I have troubles proving it so I don't know if this is the right path or not.

It's not true that $|X| = \aleph_0 \implies |X^C| = 0$. For instance, let $X$ denote the set of even numbers. Then $|X^C|$ is the set of odd numbers, so $|X| = |X^C| = \aleph_0$. Also, if you notice $C$ is just $\mathcal{P}(\mathbb{N})$, since every subset of $\mathbb{N}$ has cardinality $\leq \aleph_0$.
It's not true that $c(S) = c(C)$. Define:
$$
\mathcal{F} := \{X \subseteq \mathbb{N} : \text{$X$ is finite or $\mathbb{N} \setminus X$ is finite}\}
$$
It's not hard to check that $\mathcal{F}$ is indeed a field of sets over $\mathbb{N}$. Then since $S \subseteq \mathcal{F}$, $c(S) \subseteq \mathcal{F}$, but $\mathcal{F} \subsetneq c(C) = \mathcal{P}(\mathbb{N})$.
However, it is true that $\sigma(S) = \sigma(F) = \sigma(C)$. Can you prove it yourself?
