2 minor edits

This is an aside that I mentioned elsewhere long ago but deserves mention here since it homes in on the counterintuition that probably led Colin to doubt the answer.

As Colin pointed out, every $R \subset \omega$ can be interpreted as a binary relation on $\omega$ through a pairing function. This leads to a partition $\mathcal{B}$ of $\mathcal{P}(\omega)$ into isomorphism classes of binary relational structures $(\omega,R)$. Every countable infinite ordinal $\alpha$ has its own isomorphism class $B_\alpha \in \mathcal{B}$ and therefore $\aleph_1 \preceq \mathcal{B}$. We can also see that $2^{\omega} 2^{\aleph_0} \preceq \mathcal{B}$ in a multitude of ways. For example, we can map each $X \subseteq \omega$ to the isomorphism class of the directed graph consisting of one directed cycle of length $n+1$ for each $n \in X$ and infinitely many isolated points to fill space. In fact, we see that $\aleph_1 + 2^{\omega2^{\aleph_0} \preceq \mathcal{B}$ since the ranges of these two maps are disjoint. This is all provable without the axiom of choice.

There are models of ZF in which $2^\omega$ 2^{\aleph_0}$and$\aleph_1$are incomparable cardinals. Solovay's model where all sets of reals are Lebesgue measurable is such an example. In such models,$\mathcal{B}$must have cardinality strictly greater than$2^\omega$... 2^{\aleph_0}$... Yes, that's right: $\mathcal{B}$ is a partition of $\mathcal{P}(\omega)$ that has more pieces than there are elements in $\mathcal{P}(\omega)$!

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This is an aside that I mentioned elsewhere long ago but deserves mention here since it homes in on the counterintuition that led Colin to doubt the answer.

As Colin pointed out, every $R \subset \omega$ can be interpreted as a binary relation on $\omega$ through a pairing function. This leads to a partition $\mathcal{B}$ of $\mathcal{P}(\omega)$ into isomorphism classes of binary relational structures $(\omega,R)$. Every countable infinite ordinal $\alpha$ has its own isomorphism class $B_\alpha \in \mathcal{B}$ and therefore $\aleph_1 \preceq \mathcal{B}$. We can also see that $2^{\omega} \preceq \mathcal{B}$ in a multitude of ways. For example, we can map each $X \subseteq \omega$ to the isomorphism class of the directed graph consisting of one directed cycle of length $n+1$ for each $n \in X$ and infinitely many isolated points to fill space. In fact, we see that $\aleph_1 + 2^{\omega} \preceq \mathcal{B}$ since the ranges of these two maps are disjoint. This is all provable without the axiom of choice.

There are models of ZF in which $2^\omega$ and $\aleph_1$ are incomparable cardinals. Solovay's model where all sets of reals are Lebesgue measurable is such an example. In such models, $\mathcal{B}$ must have cardinality strictly greater than $2^\omega$... Yes, that's right: $\mathcal{B}$ is a partition of $\mathcal{P}(\omega)$ that has more pieces than there are elements in $\mathcal{P}(\omega)$!