I think proposition XIII.1.22 of Odifreddi is false but I wanted to check I wasn't being dumb. Here's the claim.


**Definition**: The set$^1$ $A$ is arithmetically-hyperimmune-free if every function $f$ arithmetic in $A$ is majorized by some arithmetic function (i.e. there is an arithmetic function $g$ s.t. $\forall x(f(x) \leq g(x))$).


XIII.1.22: The set of arithmetically-hyperimmune-free degrees is comeager. In other words, the set of $X$ such that there is some $f \leq\_a X$ not majorized by any arithmetic function is meager.


However, it seems easy to prove the claim is false. Define the arithmetic functional $f^{A}(n)$ to be the $n+1$ least element in $A$. $f^{A}$ is not only arithmetic but outright computable in $A$. Define $U\_i$ to be the union of $[\sigma]$ such that for some $n > i$, $f^{\sigma}(n)$ is defined and greater than $h\_i(n)$, the $i$-th arithmetic function. Clearly $U\_i$ is open and to see that it's dense note that if $\lvert{x : \tau(x) = 1 }\rvert = m$, and $h\_i(m+1)=k$ then $f^{\tau\hat{}0^{k+1}\hat{}1^i}(m+1)\geq k+ 1 > h\_i(m+1)$ and $f^{\tau\hat{}0^{k+1}\hat{}1^i}$ is defined on all arguments up to $i$.


But $\cap\_{i \in \omega} U\_i$ is a countable intersection of open dense sets and thus comeager and if $A \in \cap\_{i \in \omega} U\_i$ then $f^{A}$ is total and isn't majorized by any arithmetic function.


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Moreover, trying to go through the proof in Odifreddi I run into two big issues. First, the very first bullet point in the proof says that for $\psi$ arithmetical.


* { $A$ : $A \models \psi$ } is either meager of comeager


But that seems obviously false, e.g., what if $\psi$ is $A(0) = 1$. If $X$ = { $A$ : $A(0) = 1$ } then there is a homemorphism of the space mapping $X$ to it's compliment so it can't be either meager or comeager.


Second, the paper it cites to "Measure-Theoretic Uniformity in Recursion Theory and Set Theory" by Sacks seems to be all about measure (not category). It proves that for a given two place arithmetic predicate $\phi^{X}(x,y)$ the set of reals $X$ such that the reduction is either partial (for some $x$ there is no $y$) or there is an arithmetic function $f(x)$ bounding $y$. But that seems like a very different claim to me.


What am I missing here? Am I being dumb?


1: Technically the definition given says that an arithmetic degree $a$ is arithmetically-hyperimmune-free if every $A \in a$ is arithmetically-hyperimmune-free as above but that's the same thing since an arithmetic degree is just the equivalence class induced by arithmetic reducibility.



Arithmetically-hyperimmune-free degrees are comeager
Absent any pushback and based on some further reading I'm going to say this claim is false and the correct claim is that the arithmetically-hyperimmune-free degrees have measure 0.


