# A continuous map relating co-constructible reals

My question is the following:

• Given $$x,y \in \omega^\omega$$ such that $$x\equiv_c y$$ is there an $$L$$-definable continuous map $$\varphi: \omega^\omega\rightarrow \omega^\omega$$ such that $$\varphi(x) = y$$?

By $$x\equiv_c y$$ I mean that they are in the same constructibility degree, i.e. $$L(x) = L(y)$$ and by $$\varphi$$ being $$L$$-definable I mean that the map $$\varphi$$ is defined by a formula with constructible parameters. Also I can assume (if it is of any help) $$V=L(x)$$.

At first look I would say that it is unlikely, but I cannot say why.
Do you have any idea or suggestion?

Thanks!

• Might be relevant: if you add a Sacks real $s$ over $L$, then in $L(s)$, for every real $x,y$, there is a continuous map coded in $L$ that maps $x$ to $y$. This shows up as Lemma 74 in On Sacks Forcing and the Sacks Property by Stefan Geschke and Sandra Quickert. Jun 18 at 7:13

By "formula" I will mean "formula of set theory with ordinal parameters." Note that "ordinal parameters" is equivalent to "constructible parameters" for our purposes, since $$L$$ carries a definable bijection with $$\mathsf{Ord}$$.

Also, since the answer to your question is trivially true if $$V=L$$, I'm interpreting your question as asking whether we always have the situation you describe. Under this interpretation we get a highly robust negative answer: every $$M\models\mathsf{ZFC}$$ has a forcing extension in which the principle you ask about fails. So your intuition is correct in a very strong way.

We always have $$x\equiv_c x'$$.

Proof: since $$x'\ge_Tx$$ we have $$x'\ge_cx$$, and conversely since $$x'$$ is $$\Sigma^0_1(x)$$ we have $$x\ge_cx'$$. $$\quad\Box$$

Given countably many continuous functions $$f_i:\omega^\omega\rightarrow\omega^\omega$$ ($$i\in\omega$$), there is a real $$a$$ such that whenever $$x\ge_Ta$$ we have $$f_i(x)\le_Tx$$ for each $$i\in\omega$$ - and consequently $$f_i(x)\not=x'$$ for any $$i\in\omega$$.

Proof: basically, have $$a$$ "code" all the $$f_i$$s in some appropriate way. $$\quad\Box$$

Now given a (transitive, for simplicity) model $$M$$ of $$\mathsf{ZFC}$$, let $$\mathsf{ODC}(M)$$ ("ordinal-definable continuous") be the set of continuous functions $$f:\omega^\omega\rightarrow\omega^\omega$$ which are definable over $$M$$ by a formula with ordinal parameters.

If $$\mathsf{ODC}(M)$$ is countable in $$M$$, then $$M\models$$ "There is a counterexample to your question."

Proof: apply the two simple observations above, with $$y=x'$$. $$\quad\Box$$

Of course, on the face of it $$\mathsf{ODC}(M)$$ could be quite large (e.g. size continuum$$^M$$). Fortunately, we can control it relatively easily:

If $$\mathbb{P}\in M$$ is a homogeneous forcing and $$G$$ is $$\mathbb{P}$$-generic over $$M$$, then "$$\mathsf{ODC}(M)=\mathsf{ODC}(M[G])$$" - or, more precisely, if $$f\in\mathsf{ODC}(M[G])$$ then $$f\upharpoonright M\in\mathsf{ODC}(M)$$.

Proof: this is the usual homogeneity argument: the formulas defining elements of $$\mathsf{ODC}$$ only involve parameters from the ground model, so their behavior on ground model inputs is independent of the choice of generic. Now use definability of forcing. $$\quad\Box$$

The point is that if $$M,M[G]$$ are as above, then $$M[G]\models\vert\mathsf{ODC}(M[G])\vert=\vert\mathsf{ODC}(M)\vert.$$ And now to wrap up we just note that $$Col(\omega,\kappa)^M$$ is homogeneous for any $$\kappa$$, including $$\kappa=\vert\mathsf{ODC}(M)\vert^M$$.