I am facing difficulty in solving this, where I have to prove or find the counterexample for the below statement. Could anyone guide me on how do I go about solving this. Thanks in advance.
If R is a relation on a given set A such that RoRoR is a subset of R, then RoRoRoRoR is also a subset of R, where o denotes composition of the relations.
Recall that elements of $R^n$ are pairs $(a,c)$ with $a,c\in A$ such that there exist elements $b_1,b_2,b_3,\dots,b_{n-1}$ from $A$ such that $(a,b_1),(b_1,b_2),(b_2,b_3),\dots,(b_{n-2},b_{n-1}),(b_{n-1},c)$ are all elements of $R$.
Suppose that $R^3\subseteq R$. That means in particular that if we have elements $a,b_1,b_2,c$ such that $(a,b_1),(b_1,b_2),(b_2,c)$ are all elements of $R$ then $(a,c)$ is also an element of $R$.
Now, suppose that $(a,e)\in R^5$.
That means in particular that there exist $b_1,b_2,c,d_4$ such that $(a,b_1),(b_1,b_2),(b_2,c),(c,d_4),(d_4,e)$ are all pairs in $R$.
Then we have
$\underbrace{\underbrace{(a,b_1)(b_1,b_2)(b_2,c)}_{(a,c)}(c,d_4)(d_4,e)}_{(a,e)}$
That is, since $(a,b_1)(b_1,b_2)(b_2,c)$ are all elements of $R$ we have $(a,c)\in R^3$ and since $R^3\subseteq R$ we have that $(a,c)$ is in $R$ and since $(a,c)(c,d_4)(d_4,e)$ are all elements of $R$ that $(a,e)$ is also in $R^3$ implying that $(a,e)$ is also in $R$ as well.
So, as a result, $(a,e)\in R^5$ implies that $(a,e)\in R$ implying $R^5\subseteq R$