Assuming the axiom of choice, every set can be linearly (indeed, well-) ordered. However, in lieu of choice this can fail, as witnessed most drastically by the consistency of [amorphous sets]().

I'm interested in what the situation with the Turing degrees can be, if we assume AD. That is:

> In ZF+DC+AD *(plus whatever else is needed to get a good answer)*, can the set $\mathcal{D}$ of Turing degrees be linearly ordered?

(Note that of course there is no demand that this ordering behave nicely with respect to $\le_T$, in any sense.)

I strongly suspect the answer is "no," but I don't immediately see how to prove it.

In fact, as far as I can tell very few sets of Turing degrees admit "definable" linear orderings - specifically, every example I can find is the image of some injective partial function $f$: $\subseteq2^\omega\rightarrow\mathcal{D}$. For example, the standard construction of a continuum-sized antichain of Turing degrees consists of building a continuous function $g:2^\omega\rightarrow 2^\omega$ such that $x\not=y\implies deg(g(x))\not=deg(g(y))$; we then push the lexicographic order on $2^\omega$ through the map $f=deg\circ g$. This raises the following question, especially assuming a negative answer to (1):

> Is there a reasonable extension of ZF which proves that every orderable set of Turing degrees is the image of some injective function from a subset of $2^\omega$?