The free Boolean algebra on a denumerable set of free generators is atomless and denumerable. In fact, it is up to isomorphic copies the only atomless denumerable Boolean algebra.
Consider now the free (Boolean) $\sigma$-algebra on a denumerable set of free generators. Is it also atomless? What is its cardinality?
Suppose $X$ is the free (Boolean) $\sigma$-algebra on generators $e_1, e_2, \ldots$. Then first, we claim: for any $S \subseteq \mathbb{N}^+$, let $x_n^S := \begin{cases} e_n, & n \in S\\e_n^c, & n \notin S, \end{cases}$ and $x^S := \bigwedge_{n=1}^\infty x_n^S$. (So for example, $x^{\{ 1, 3, 5, \ldots \}} = e_1 \wedge e_2^c \wedge e_3 \wedge e_4^c \wedge e_5 \wedge \cdots$.) Then the collection of $x^S$ for $S \subseteq \mathbb{N}^+$ is a collection of $\mathfrak{c}$ distinct atoms of $X$.
To see this, observe that $\{ y \in X \mid (x_n^S \le y) \vee (x_n^S \wedge y = \bot) \}$ is a sub-$\sigma$-algebra of $X$, and each $e_m$ is in this set. Since $X$ must be generated as a $\sigma$-algebra by $e_1, e_2, \ldots$, it follows that this set is all of $X$, showing that $x_n^S$ is an atom of $X$.
To show they are all distinct, consider the $\sigma$-algebra $\mathcal{B}$ contained in $P(P(\mathbb{N}^+))$ and generated by $A_n := \{ S \in P(\mathbb{N}^+) \mid n \in S \}$. Then under the unique morphism $X \to \mathcal{B}$ sending $e_n \mapsto A_n$, the image of $x_n^S$ is exactly $S$.
As for the cardinality: the standard argument that there are $\mathfrak{c}$ Borel subsets of $\mathbb{R}$ can be adapted to show $|X| = \mathfrak{c}$. Namely, define $X_0 := \{ e_1, e_2, \ldots \}$, and by transfinite recursion, define $X_{\alpha+1}$ to be the set of countable unions, countable intersections, or complements of elements of $X_\alpha$, and $X_\beta := \bigcup_{\alpha < \beta} X_\alpha$ for limit ordinals $\beta > 0$. Then by transfinite induction, $|X_\alpha| \le \mathfrak{c}$ for each $\alpha \le \omega_1$; and since $X_{\omega_1}$ is a sub-$\sigma$-algebra of $X$ containing each $e_n$, then $X = X_{\omega_1}$. Therefore, $|X| \le \mathfrak{c}$. On the other hand, since we previously showed $X$ has $\mathfrak{c}$ distinct atoms, $|X| \ge \mathfrak{c}$.
I would conjecture that the $\sigma$-algebra $\mathcal{B}$ mentioned in the proof, which is the set of Borel subsets of $P(\mathbb{N}^+)$ (where we give the product topology to this space), is in fact isomorphic to $X$ via the morphism with $e_n \mapsto A_n$, i.e. the sets $A_n$ are independent generators. I don't currently have a proof of this conjecture, however.