complexity - Non Polynomial reduction I have a question, does non-polynomial reduction exists for all problems?
the issue is only regarding polynomial reductions?
You can show that it is not the case that every two languages are reducible to each other using a counting argument. Consider the class of all  languages $ 2^{\Sigma^*}$. If by contradiction every two languages are reducible to each other, then every language is an $2^{\Sigma^*}$-hard language (in particular, there is a language that all languages reduce to). One can show that there are more languages than reductions. Thus, an $2^{\Sigma^*}$-hard language
must have two different languages that reduce to it using the same reduction, but this is impossible.
More formally, assume by contradiction that there is an $2^{\Sigma^*}$-hard language $L$. We know that there are $2^{\aleph_0}$ languages, yet, as there are only $\aleph_0$ TMs, there are also only $\aleph_0$ reductions. Then, it follows that there are distinct languages $A$ and $B$ such that
$A \leq_m L$ and $B \leq_m L$, but the reductions from $A$ and $B$ to $L$ coincide. Denote this reduction
by $f$. Since $A\neq B$, there exists w.l.o.g a word $x\in A\setminus B$. Since $f$ is a reduction from $A$ to
$L$, we get that $f(x) ∈ L$. But $f$ is also a reduction from $B$ to $L$, and thus we get that $f(x) \notin L$, and we
reached a contradiction.