Let $R$ be any relation over $\mathbb Z$ (the integers). Let $R_n $ be the set of all relations $ R $ (over the integers) such that $n$ is natural, by:
$ R_0:= R$.
$R_{n+1}:= R_n \cup (R_n ◦ R_n)$.
$R_∞=\bigcup\limits_{n=0}^{\infty} R_{n}$.
(couldn't figure out how to write the index correctly so i just wrote it regularly).
Show that for every natural number $n$, exist a relation $ R$ (over the integers) such that; $R_{n+1}=R_∞ $ and $R_n≠R_∞$
Each stage makes it so that if $xR_ny$ and $yR_nz$ then $xR_{n+1}z$. Intuitively, you can think of this as $(x,z)$ being in the relation $R_{n+1}$ if there is a path of lenth $\le 2$ between $x$ and $z$ in $R_n$. Or length $\le 4$ in $R_{n-1}$, length $\le 8$ in $R_{n-2}$ etc. So $(x,z)$ is in $R_n$ if there is a path of length $\le 2^n$ in $R_0$.
So what this comes down to is finding a relation where all $x$ and $z$ that can be connected by a path are connected by a path of length $2^{n+1}$, but some of them need a path longer then $2^n$. Can you think of a relation like that for a given $n$?
You also don't need the relation you first think of to be on the integers. If you make it on any set smaller than the integers, you can fit it into a relation on the integers so that it has the same property.