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
- $R\subseteq S$;
- $S$ is reflexive; and
- $S$ is transitive.
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 $R$.
So far, it seems you have $R=\{(x, y)\mid y=x+1, \text{ or}\; y=x \text{ or}\; y = x-1\}$.
For transitive closure, you need to have the relation satisfy, for all $x, y, z$, if $(x, y) \in R$ and $(y, z) \in R$, then your need to ensure $(x, z) \in R$.
Test your current conditions for the relation and check for transitivity.
For example, if $y = x + 1$ and $z = y + 1$, then $z = (x + 1) + 1 \implies z = x+2$. So currently, you don't have transitive closure. Both $(x, y), (y, z)\in R$, but $(x,z) \notin R$.
Note that $R = \{(x, y) \mid y = x\}$, by itself, is an equivalence relation (hence reflexive, symmetric, and transitive for all elements on which it is defined), as equality is perhaps the most fundamental of all equivalence relations.
Assuming we are talking about real numbers, we can get transitive closure (with reflexive closure), on the reals using the relation $R = \{(x, y)\mid y\leq x\}$. But this relation fails to by symmetric. "$\leq$" is a paradigmatic partial order relation on the set of reals: reflexive, antisymmetric, and, and transitive. Can you see why it is transitive, reflexive, but not symmetric?
Best Answer
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?