Let $R=\{\langle 1,2\rangle,\langle 2,1\rangle,\langle 2,3\rangle,\langle 3,4\rangle,\langle 4,1\rangle\}$ on $\{1,2,3,4\}$. What failure of transitivity do we have here?
$$\begin{align*}
&\langle 1,2\rangle,\langle 2,1\rangle\in R,\text{ but }\langle 1,1\rangle\notin R\\
&\langle 1,2\rangle,\langle 2,3\rangle\in R,\text{ but }\langle 1,3\rangle\notin R\\
&\langle 2,1\rangle,\langle 1,2\rangle\in R,\text{ but }\langle 2,2\rangle\notin R\\
&\langle 2,3\rangle,\langle 3,4\rangle\in R,\text{ but }\langle 2,4\rangle\notin R\\
&\langle 3,4\rangle,\langle 4,1\rangle\in R,\text{ but }\langle 3,1\rangle\notin R\\
&\langle 4,1\rangle,\langle 1,2\rangle\in R,\text{ but }\langle 4,2\rangle\notin R\\
\end{align*}$$
This means that in order to expand $R$ to a transitive relation, we must add at least the six ordered pairs $\langle 1,1\rangle,\langle 1,3\rangle,\langle 2,2\rangle,\langle 2,4\rangle,\langle 3,1\rangle$, and $\langle 4,2\rangle$. This gives us the relation
$$R\,'=\{\langle 1,1\rangle,\langle 1,2\rangle,\langle 1,3\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,1\rangle,\langle 3,4\rangle,\langle 4,1\rangle,\langle 4,2\rangle\}\;.$$
Are there any failures of transitivity here?
$$\begin{align*}
&\langle 1,2\rangle,\langle 2,4\rangle\in R\,',\text{ but }\langle 1,4\rangle\notin R\,'\\
&\langle 1,3\rangle,\langle 3,4\rangle\in R\,',\text{ but }\langle 1,4\rangle\notin R\,'\\
&\langle 3,1\rangle,\langle 1,3\rangle\in R\,',\text{ but }\langle 3,3\rangle\notin R\,'\\
&\langle 3,1\rangle,\langle 1,2\rangle\in R\,',\text{ but }\langle 3,2\rangle\notin R\,'\\
&\langle 3,4\rangle,\langle 4,2\rangle\in R\,',\text{ but }\langle 3,2\rangle\notin R\,'
\end{align*}$$
Thus, $R\,'$ isn’t yet transitive: we need to add at least the pairs $\langle 1,4\rangle,\langle 3,2\rangle$, and $\langle 3,3\rangle$ to $R\,'$ to have any hope of having a transitive relation. This gives us the relation
$$R''=\{\langle 1,1\rangle,\langle 1,2\rangle,\langle 1,3\rangle,\langle 1,4\rangle,\langle 2,1\rangle,\langle 2,2\rangle,\langle 2,3\rangle,\langle 2,4\rangle,\langle 3,1\rangle,\langle 3,2\rangle,\langle 3,3\rangle,\langle 3,4\rangle,\langle 4,1\rangle,\langle 4,2\rangle\}\;.$$
Note that in $R''$ the elements $1,2$, and $3$ of $A$ are all related to everything in $A$; only $4$ is not. And that gives us some failures of transitivity in $R''$:
$$\begin{align*}
&\langle 4,1\rangle,\langle 1,3\rangle\in R'',\text{ but }\langle 4,2\rangle\notin R''\\
&\langle 4,1\rangle,\langle 1,4\rangle\in R'',\text{ but }\langle 4,4\rangle\notin R''
\end{align*}$$
There are some other failures, but in each case the missing pair is either $\langle 4,3\rangle$ or $\langle 4,4\rangle$. Thus, to make $R''$ transitive we must at least add these two pairs. Call the resulting relation $\overline{R}$. We can’t add any more pairs, because $\overline{R}=A\times A$: it already contains every possible ordered pair. And finally we do have a transitive relation.
Because at each stage I added only those ordered pairs that that I had actually shown to be necessary for transitivity, this $\overline{R}$ is the smallest transitive relation containing the original relation $R$; by definition it is the transitive closure of $R$. The problem asks you to find and write out the transitive closures of the other three relations as well.
There are much more efficient ways to find the transitive closure of a relation, but at this point it’s probably most instructive for you to do everything by hand, so to speak, as I did above.
Heres a nice theorem which helps in determining whether or not a relation is transitive. Let's define a relation $\mathbf{R}$ on a set $A$. We know that this defines a relation $\mathbf{R}:A \longrightarrow A$. Now, we want to prove that $\mathbf{R}$ is a transitive relation.
Theorem: $\mathbf{R}$ is a transitive relation if and only if $\mathbf{R}^{n} \subseteq \mathbf{R}$ for every positive integer $n$.
$\mathbf{Proof} (\longrightarrow)$:(via mathematical induction)
Suppose $\mathbf{R}$ is a transitive relation. Define the propositional function $P(n)$ such that $P(n)$ implies that $\mathbf{R}^{n} \subseteq \mathbf{R}$. Now we check the base case: $n=1$. Since $\mathbf{R}^{1}=\mathbf{R}$, it follows that $\mathbf{R}^{1} \subseteq \mathbf{R}$ and so $P(1)$ is true. We now must show that if $P(n)$ is true, then $P(n+1)$ is true. In other words, if $\mathbf{R}^{n} \subseteq \mathbf{R}$ then $\mathbf{R}^{n+1} \subseteq \mathbf{R}$. Let $\alpha,\beta \in A$ such that $\alpha$ and $\beta$ are related under $\mathbf{R}^{n+1}$. More compactly, we may write this as $(\alpha, \beta) \in \mathbf{R}^{n+1}$. Recall that $\mathbf{R}^{n+1}$ is the composition of $\mathbf{R}^{n}$ and $\mathbf{R}$; in other words: $\mathbf{R}^{n+1} = \mathbf{R}^{n} \circ \mathbf{R}$. Now, since $(\alpha,\beta) \in \mathbf{R}^{n+1}$, it follows from the definition of a transitive relation that there exists $\theta \in A$ such that $(\alpha,\theta) \in \mathbf{R}$ and $(\theta, \beta) \in \mathbf{R}^{n}$. Since $\mathbf{R}^{n} \subseteq \mathbf{R}$, we know that $(\theta, \beta) \in \mathbf{R}$. Since we defined $\mathbf{R}$ as a transitive relation, it follows that $(\alpha ,\beta) \in \mathbf{R}$. Therefore $\mathbf{R}^{n+1} \subseteq \mathbf{R}$.
$\mathbf{Proof} (\longleftarrow)$:
Let $a,b,c \in \mathbf{R}$ such that $(a,b) \in \mathbf{R}$ and $(b,c) \in \mathbf{R}$. Now can compose $\mathbf{R}$ with itself; in other words, take $\mathbf{R} \circ \mathbf{R}$. If we do so, then $(a,c) \in \mathbf{R} \circ \mathbf{R} = \mathbf{R}^{2}$. By hypothesis,$\mathbf{R}^{2} \subseteq \mathbf{R}$, and thus $(a,c) \in \mathbf{R}$. Therefore $\mathbf{R}$ is transitive.
This completes the proof.
Now you have posed a very insightful question regarding the number of transitive relations on a set. Until now, the only way to compute the number of transitive relations is to find them via brute-force methodology: counting transitive relations is an open problem in combinatorics!
Hope this helped.
Best Answer
Take each of the $n$ elements of $A$ one by one and decide whether to put it in $B$ or in $C$. No element can go in both $B$ and $C$ because they are supposed to be disjoint, so there are three options: in $B$, or in $C$, or neither. The total number of pairs is $3^n$.