Is every sigma-algebra of cardinality $2^\kappa$ for some cardinal $\kappa$? I know that finite sigma-algebras have cardinality of the form $2^n$ for some integer $n$ and that any infinite sigma-algebra is at least of cardinality $2^{\aleph_0}$.
This suggests that maybe every sigma-algebra is of cardinality $2^\kappa$ for some cardinal $\kappa$. Is this true? If so, why?
 A: Not necessarily.
Suppose that $2^{\aleph_0} = \aleph_1$ and $2^{\aleph_1}=\aleph_3$. It is still the case that $\aleph_2^{\aleph_0} = \aleph_2$.
Now take a collection of $\aleph_2$ subsets of $\omega_1$ and let them generate a $\sigma$-algebra. It will have size $\aleph_2$ which is not a cardinal of any power set. 
A: I think the following counterexample works in ZFC.
Set $\lambda = \beth_{\omega_1}$, so that $\lambda = \bigcup_{\alpha < \omega_1} \beth_\alpha$. Consider the $\sigma$-algebra $\mathcal{F}$ consisting of all countable and co-countable subsets of $\lambda$.  ("Countable" here will mean "finite or countably infinite").
I claim $|\mathcal{F}| = \lambda$.  It is clear that $|\mathcal{F}| \ge \lambda$ because $\mathcal{F}$ contains all the singletons.  
For the other direction, let $\mathcal{F}_0$ consist of all the countable subsets of $\lambda$, and $\mathcal{F}_1$ all the co-countable subsets.  Suppose $A \subset \lambda$ is countable, so $A = \{x_1, x_2, \dots\}$ (where I allow the sequence to be finite).  For each $n$, there exists $\alpha_n < \omega_1$ with $x_n \in \beth_{\alpha_n}$.  Letting $\alpha = \sup_n \alpha_n < \omega_1$ (since there are either finitely or countably many $n$), we have $A \subset \beth_\alpha$; that is, $A \in \mathcal{P}(\beth_{\alpha})$.  So, $\mathcal{F}_0 \subset \bigcup_{\alpha < \omega_1} \mathcal{P}(\beth_\alpha)$.  But $\mathcal{P}(\beth_\alpha)$ by definition has cardinality $\beth_{\alpha + 1} < \lambda$.  Thus, $\mathcal{F}$ is contained in a union of $\omega_1$ sets each having cardinality less than $\lambda$, so $|\mathcal{F}_0| \le \omega_1 \cdot \lambda = \lambda$ (since $\lambda > \omega_1$).  There is clearly a bijection between $\mathcal{F}_0$ and $\mathcal{F}_1$, so $\mathcal{F} \le \lambda + \lambda = \lambda$.
On the other hand, $\lambda$ is not of the form $2^\kappa$.  If $\kappa \ge \lambda$ then $2^\kappa > \lambda$ by Cantor's theorem.  And if $\kappa < \lambda$ then $\kappa \le \beth_\alpha$ for some $\alpha < \omega_1$, so that $2^\kappa \le 2^{\beth_\alpha} = \beth_{\alpha + 1} < \lambda$.  In either case, $2^\kappa \ne \lambda$.

Moreover, $\beth_{\omega_1}$ is consistently the least possible cardinality of a counterexample.
Proposition. Assume GCH.  Then every $\sigma$-algebra $\mathcal{F}$ with $|\mathcal{F}| < \beth_{\omega_1}$ has $|\mathcal{F}| = 2^{\kappa}$ for some $\kappa$.
Proof.  We already know it's true for $\mathcal{F}$ finite, so suppose $\mathcal{F}$ is infinite.  By GCH, every cardinal is a beth number, so $|\mathcal{F}| = \beth_\alpha$ for some $\alpha < \omega_1$.  Since we know that there are no countable $\sigma$-algebras (this can also be seen by a variant of the argument below), we have $\alpha > 0$.  I claim that $\alpha$ must be a successor ordinal; then we are done, for if $\alpha = \beta+1$ then $|\mathcal{F}| = \beth_{\beta + 1} = 2^{\beth_\beta}$.  So suppose $\alpha$ is a limit ordinal; it is countable, so $|\mathcal{F}| = \beth_\alpha$ has countable cofinality, so we can write $\mathcal{F} = \bigcup_{n=1}^\infty \mathcal{C}_n$ where $|\mathcal{C}_n| < |\mathcal{F}|$ for all $n$.  Suppose without loss of generality that every $\mathcal{C}_n$ is uncountable and that $\mathcal{C}_1 \subset \mathcal{C}_2 \subset \dots$.  Now let $\mathcal{F}_n = \sigma(\mathcal{C}_n) \subset \mathcal{F}$.  As noted by Joel David Hamkins on MathOverflow, we have $$|\mathcal{F}_n| = |\mathcal{C}_n|^{\aleph_0} \le 2^{|\mathcal{C}_n|} < |\mathcal{F}|.$$
The first inequality holds by identifying a function $f : \aleph_0 \to \mathcal{C}_n$ with a subset of $\aleph_0 \times \mathbb{C}$, where $|\aleph_0 \times \mathcal{C}_n| = |\mathcal{C}_n|$ because $\mathcal{C}_n$ is infinite.  The second inequality is because $\mathcal{F}$ is a limit beth number.
Hence we have $\mathcal{F} = \bigcup_{n=1}^\infty \mathcal{F}_n$, where $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dots$ and each is properly contained in $\mathcal{F}$.  Passing to a subsequence, we can suppose they increase strictly.  But as shown in [1], a countable strictly increasing union of $\sigma$-algebras can never be a $\sigma$-algebra, so this is a contradiction.
[1] Broughton, Allen; Huff, Barthel W., A comment on unions of sigmafields, Am. Math. Mon. 84, 553-554 (1977). ZBL0372.60004.  Thanks to hot_queen for this reference.

More generally, this can easily be extended to show that for every limit ordinal $\alpha$ of uncountable cofinality, there exists a $\sigma$-algebra of cardinality $\beth_\alpha$ which is not of the form $2^\kappa$ for any $\kappa$; and under GCH, no other cardinality is possible.
