TITLE: How many well orderings of $\aleph_0$ are there?
QUESTION [8 upvotes]: What is known about the set of well orderings of $\aleph_0$ in set theory without choice?  I do not mean the set of countable well-order types, but the set of all subsets of $\aleph_0$ which (relative to a pairing function) code well orderings.  And I would be interested in an answer in, say, ZF without choice.  My actual concern is higher order arithmetic.
I would not be surprised if ZF proves there are continuum many.  But I don't know.
At the opposite extreme, is it provable in ZF that there are not more well orderings of $\aleph_0$ than there are countable well-order types?

REPLY [7 votes]: 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 $2^{\aleph_0} \preceq \mathcal{B}$ 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 $\aleph_1 + 2^{\aleph_0} \preceq \mathcal{B}$ 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 $2^{\aleph_0}$ 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 $2^{\aleph_0}$... Yes, that's right: $\mathcal{B}$ is a partition of $\mathcal{P}(\omega)$ that has more pieces than there are elements in $\mathcal{P}(\omega)$!