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Fixed to indicate that this doesn't solve the question.

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edited Sep 22, 2017 at 18:27

Yes,This doesn't answer the second part of MC contradicts KC. More stronglyquestion, ifbut it cannot be that KC is true and $\leq_m$ is a prewellordering for justall the Borel Turing invariant functionswhich are not constant a cone (not just the increasing ones). This is because as we show below, thenthis implies there is a $\Delta^1_2$ wellordering of $\mathbb{R}$. This contradicts (contradicting AD. Assuming AC it also contradicts the existence of large or AC+large cardinals since it implies that $\mathbb{R} = \mathbb{R}^{L[x]}$ for some real $x$ by a result of Mansfield).

We prove the claim: suppose $\equiv_T$ was a universal countable Borel equivalence relation, let $=_\mathbb{R}$ be the equality relation on $\mathbb{R}$, and consider the relation $=_\mathbb{R} \times \equiv_T$ which must be Borel reducible to $\equiv_T$ via some Borel reduction $f$. For each $x \in \mathbb{R}$, let $f_x$ be the Borel Turing invariant function where $f_x(y) = f((x,y))$. IfThese $f_x$ are not constant one a cone and if $x \neq x'$, then for all $y$, $f_x(y) \not \equiv_T f_{x'}(y)$ since(because $f$ is a Borel reduction). Hence, the functions $f_x$ are all distinct under $\leq_m$ which wellorders them. So the ordering $x \leq x'$ iff $f_x \leq_m f_{x'}$ is a $\Delta^1_2$ wellordering of $\mathbb{R}$.

Yes, the second part of MC contradicts KC. More strongly, if KC is true and $\leq_m$ is a prewellordering for just the Borel Turing invariant functions, then there is a $\Delta^1_2$ wellordering of $\mathbb{R}$. This contradicts AD. Assuming AC it also contradicts the existence of large cardinals since it implies that $\mathbb{R} = \mathbb{R}^{L[x]}$ for some real $x$ by a result of Mansfield.

We prove the claim: suppose $\equiv_T$ was a universal countable Borel equivalence relation, let $=_\mathbb{R}$ be the equality relation on $\mathbb{R}$, and consider the relation $=_\mathbb{R} \times \equiv_T$ which must be Borel reducible to $\equiv_T$ via some Borel reduction $f$. For each $x \in \mathbb{R}$, let $f_x$ be the Borel Turing invariant function where $f_x(y) = f((x,y))$. If $x \neq x'$, then for all $y$, $f_x(y) \not \equiv_T f_{x'}(y)$ since $f$ is a Borel reduction. Hence, the functions $f_x$ are all distinct under $\leq_m$ which wellorders them. So the ordering $x \leq x'$ iff $f_x \leq_m f_{x'}$ is a $\Delta^1_2$ wellordering of $\mathbb{R}$.

This doesn't answer the question, but it cannot be that KC is true and $\leq_m$ is a prewellordering for all the Borel Turing invariant which are not constant a cone (not just the increasing ones). This is because as we show below, this implies there is a $\Delta^1_2$ wellordering of $\mathbb{R}$ (contradicting AD or AC+large cardinals).

We prove the claim: suppose $\equiv_T$ was a universal countable Borel equivalence relation, let $=_\mathbb{R}$ be the equality relation on $\mathbb{R}$, and consider the relation $=_\mathbb{R} \times \equiv_T$ which must be Borel reducible to $\equiv_T$ via some Borel reduction $f$. For each $x \in \mathbb{R}$, let $f_x$ be the Borel Turing invariant function where $f_x(y) = f((x,y))$. These $f_x$ are not constant one a cone and if $x \neq x'$, then for all $y$, $f_x(y) \not \equiv_T f_{x'}(y)$ (because $f$ is a Borel reduction). Hence, the functions $f_x$ are all distinct under $\leq_m$ which wellorders them. So the ordering $x \leq x'$ iff $f_x \leq_m f_{x'}$ is a $\Delta^1_2$ wellordering of $\mathbb{R}$.

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created Sep 22, 2017 at 16:04

cone

We prove the claim: suppose $\equiv_T$ was a universal countable Borel equivalence relation, let $=_\mathbb{R}$ be the equality relation on $\mathbb{R}$, and consider the relation $=_\mathbb{R} \times \equiv_T$ which must be Borel reducible to $\equiv_T$ via some Borel reduction $f$. For each $x \in \mathbb{R}$, let $f_x$ be the Borel Turing invariant function where $f_x(y) = f((x,y))$. If $x \neq x'$, then for all $y$, $f_x(y) \not \equiv_T f_{x'}(y)$ since $f$ is a Borel reduction. Hence, the functions $f_x$ are all distinct under $\leq_m$ which wellorders them. So the ordering $x \leq x'$ iff $f_x \leq_m f_{x'}$ is a $\Delta^1_2$ wellordering of $\mathbb{R}$.