thought it would be necessary to actually write a proof for this, since it's not on stack exchange and the only proof i could find was written by the dude himself here
but just to make sure we're on the same page here, i'm going to prove the following:
Let $A \subseteq \mathbb{N}$ and $A \ge_T K$ where $K$ is the standard halting problem.
Then, there exists some other $B \subseteq \mathbb{N}$ s.t. $A \equiv_T B'$, where $B'$ denotes the Turing jump of $B$, so all relations that can be decided by the halting problem relativied to $B$.
hope i haven't used too many nonstandard things here, from what i've seen there's been like a bajillion ways to format this stuff unlike with analysis
just in case people are still confused, all notation i use comes from these notes
the idea is to start with $A$ (and $K$ by extension) and try to build a characteristic function that both:
which seems pretty hard, and... well, yeah, it highkey is.
but using the finite extension method, we can construct
$\sigma_0 \sqsubset \sigma_1 \sqsubset \sigma_2 \sqsubset \cdots$
and make their limit $\text{Char}_B$
(our base case is just $\sigma_0$ being the empty sequence)
at even stages, we're trying to construct $\sigma_{s+1}$ where $s=2e$.
to do this, we ask whether there exists a $\sigma \sqsupset \sigma_s$ s.t. $\varphi_e^\sigma(e)\downarrow$,
where $\varphi_e$ denotes the $e$-th program in some enumeration of all recursive programs
(notice that this question can be solved by the halting problem, which we get with $A \ge_T K$)
if there is, we take the first such $\sigma$ and set that as our $\sigma_{s+1}$. given this extension, we know that using $A$, gives us $\varphi_e^B(e)\downarrow$ for this specific $e$
if not, we do nothing and let $\sigma_{s+1}=\sigma_s$. no matter how we extend/complete $B$ in the future, $\varphi_e^B(e)\uparrow$
now by this construction, we've forced $A \ge_T B'$, the halting problem relativied to $B$. a procedure could literally just repeat this construction and check the result for the question that was asked when $s=2e$.
ok so before i do this notice that since we're using $A$, we can enumerate its elements in increasing order like $a_0 < a_1 < a_2 < \cdots$
so if we're trying to construct $\sigma_{s+1}$ where $s=2e+1$, ok lol this is gonna be really stupid:
\[\sigma_{s+1}=\sigma_s^\frown (0, \underbrace{1, 1, \cdots, 1}_{a_e\text{ ones}}, 0)\](the weird curve on top means sequence conatentation btw)
now this allows $B' \ge_T A$, since now using $B'$ we can enumerate the elements of $A$ in increasing order and by extension decide $A$
why is this? well, notice that since we have $B'$, we also have $K$! thus, we can simulate the construction in even stages just like we did using $A$
then, after each even stage, we look at how many $1$s are jammed between the $0$s, which i'm like 90% sure is a computable procedure. this will give us the elements of $A$ in increasing order.
can i say $\square$ now? i'm just gonna say it. a crappy proof is better than no proof, ig.