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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 start with two simple observations:

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). 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 $\kappa=\vert\mathsf{ODC}(M)\vert^M$.

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 start with two simple observations:

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 $\kappa=\vert\mathsf{ODC}(M)\vert^M$.

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 start with two simple observations:

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). 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. $\quad\Box$

And at this point we just note that $Col(\omega,\kappa)^M$ is homogeneous for any $\kappa$, including $\kappa=\vert\mathsf{ODC}(M)\vert^M$.

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 start with two simple observations:

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). 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 $\kappa=\vert\mathsf{ODC}(M)\vert^M$.

There's some flexibility around your notion of "$L$-definable" - is the formula $\varphi$ allowed to quantify over all of $V$, or is it really a continuous function with code in $L$?

We get the same answer in each case, but a bit more care is needed in the first interpretation.

Easy version

If we look at continuous functions with constructible codes, then the existence of such a $\varphi$ is equivalent to a Turing reduction from $y$ to $x\oplus z$ for some constructible real $z$. So we can repose the question as follows:

If $x,y$ are reals with $x\equiv_cy$, must there be some $z\in L$ with $x\oplus z\ge_T y$?

The answer is definitely no. For example, any two Turing-inequivalent but constructibility-equivalent reals $x,y$ each of which computes every constructible real will do.

Note that this gives an "internally-describable" counterexample: the Turing jump map $x\mapsto x'$ has a Borel code in $L$, so we basically have a Borel-coded-in-$L$ way to transform a witness to the countability of $\mathbb{R}^L$ into a counterexample to your principle.

Harder version

OK, what if we genuinely look at formulas with ordinal parameters which when evaluated in $V$ happen to give continuous functions?

(Note that I'm switching from "constructible parameters" to "ordinal parameters" here for simplicity; since $L$ carries a definable well-ordering of ordertype $\mathsf{Ord}$, this isn't an issue.)

Now we need to do a bit of work. The key is that there is a forcing extension $V[G]$ of the universe such that the set of continuous functions $\mathbb{R}\rightarrow\mathbb{R}$ definable over $V[G]$ with ordinal parameters is countable. If you believe me, then we can run the codes-based argument above inside this forcing extension: in $V[G]$, there will be a real $x$ coding all the graphs of the ordinal-parameter-definable continuous functions, and we will have $x\equiv_cx'$ but $x$ and $x'$ can't be "connected" by any ordinal-parameter-definable continuous function for Turing-reducibility reasons.

The subtlety, of course, is that the same formula $\varphi$ with ordinal parameters may code different continuous functions in $V$ versus in $V[G]$. To get around this, note that homogeneous forcings don't do too much damage - if $\varphi$ (with ordinal parameters) defines a continuous function over $V[G]$, and $G$ is generic for a homogeneous forcing, then there is some formula $\psi$ (again with ordinal parameters) which defines the same function over $V$; or, more accurately, $\psi^V$ has a unique continuous extension to $\mathbb{R}^{V[G]}$, and that continuous extension is $\varphi^{V[G]}$.

And Levy collapses are homogeneous.

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 start with two simple observations:

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). 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. $\quad\Box$

And at this point we just note that $Col(\omega,\kappa)^M$ is homogeneous for any $\kappa$, including $\kappa=\vert\mathsf{ODC}(M)\vert^M$.