Many-one reductions between the set of true sentences and a particular arithmetical set
For any such $X$ we have $X\le_mTh(\mathbb{N})$ but not conversely (and in fact $Th(\mathbb{N})$ is vastly more complicated than $X$).
To see that $X\le_mTh(\mathbb{N})$, we want to translate a question of the form &quot;Is $n\in X$?&quot; into one of the form &quot;Is $\psi_n$ a true sentence?&quot; for some appropriate sentence $\psi_n$. Now &quot;$n\in X$&quot; is equivalent to &quot;$\gamma(n)$,&quot; and that's almost a sentence in the language of arithmetic - the only issue is that &quot;$n$&quot; isn't a symbol in the language of arithmetic (which consists only of $0,1,+,\times,&lt;$). Do you see a quick way around this?

 Each natural number $n$ has a corresponding numeral, usually denoted &quot;$\underline{n}$,&quot; which is just an appropriate string of the symbols $0$,&quot; &quot;$1$,&quot; &quot;$($,&quot; and &quot;$)$&quot; (or is the symbol &quot;$0$&quot; if $n$ is $0$): e.g. $\underline{4}$ is the string $((1+1)+1)+1$. Then if we let $\psi_n=\gamma(\underline{n})$, we have that $\psi_n\in Th(\mathbb{N})$ iff $n\in X$.

Note that here I'm conflating e.g. the number $0$ and the constant symbol $0$; this is annoying, but standard.

The other direction is a consequence of Tarski's undefinability theorem: just show that anything many-one reducible (or even Turing-reducible) to an arithmetically definable set is itself arithmetically definable. This will show that we never have $Th(\mathbb{N})\le_mX$ (or even $\le_TX$) for such an $X$. The key here is to show that the basic notions of computability theory are arithmetically definable:

 Use Kleene's $T$-predicate to talk about computable reductions.