The reflexive closure of $R$ is the smallest relation $S$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq S$ and (2) $S$ is reflexive. To form this, just add to $R$ the ordered pairs required by reflexivity:
$$S=R\cup\{\langle a,a\rangle,\langle b,b\rangle,\langle c,c\rangle,\langle d,d\rangle,\langle e,e\rangle\}\;.$$
Similarly, the symmetric closure $R$ is the smallest relation $T$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq T$ and (2) $T$ is symmetric. Here again you need only add to $R$ the missing ordered pairs needed for symmetric. There is only one; what is it?
Finally, the transitive closure of $R$ is the smallest relation $U$ on $\{a,b,c,d,e\}$ such that (1) $R\subseteq U$ and (2) $U$ is transitive. Finding the transitive closure can be little trickier than finding the reflexive or symmetric closure, because you may have to go through several stages of adding ordered pairs. $R$ has the ordered pairs $\langle a,c\rangle$ and $\langle c,a\rangle$, so $U$ will have those pairs as well. In order to be transitive, $U$ will therefore also have to include the pair $\langle a,a\rangle$. Similarly reasoning, reversing the rôles of $a$ and $c$, shows that $U$ must contain the ordered pair $\langle c,c\rangle$. And we can apply the same reasoning to the pairs $\langle b,d\rangle,\langle d,b\rangle\in R$ to conclude that $U$ must contain $\langle b,b\rangle$ and $\langle d,d\rangle$. Is there anything else? Yes: $R$ includes both $\langle e,d\rangle$ and $\langle d,b\rangle$, so $U$ does as well and must therefore contain $\langle e,b\rangle$ in order to be transitive. At this point we know that we must have at least
$$R\cup\{\langle a,a\rangle,\langle b,b\rangle,\langle c,c\rangle,\langle d,d\rangle,\langle e,b\rangle\}\;.\tag{1}$$
Is the relation $(1)$ transitive? If so, it’s the transitive closure of $R$. If not, what do you need to add to it to make it transitive?
Best Answer
Suppose that $R$ is a relation on a set $A$. The reflexive transitive closure of $R$ is the smallest relation $S$ on $A$ such that
If $R$ is already reflexive and transitive, then $R$ is its own reflexive transitive closure, but that’s not the case with your covering relations.
One way of constructing the reflexive transitive closure of $R$ is to begin by expanding $R$ to $$R_r=R\cup\{\langle a,a\rangle:a\in A\}\;,$$ adding to $R$ all of the pairs $\langle a,a\rangle$ that aren’t already in it. Then whenever you have $\langle x,y\rangle$ and $\langle y,z\rangle$ in $R_r$, you throw in $\langle x,z\rangle$ if it’s not already there to get $R_r^2$. Repeat to get $R_r^3$. If $A$ is finite, after a finite number of steps nothing new will be added; the resulting relation $R_r^*$ is the reflexive transitive closure of $R$.