A function is well defined, that means $x=y\implies f(x)=f(y)$ but you have once $f(3)=1$ and $f(3)=3$
try for the rst. for your control, b is no function, d is no function but e is a function.
In b) you have no function because $f(3)$ does not exist, but a function assigns to every object of your domain, an object of your codomain.
Lets recall the definition of transitivity. A relation $R \subseteq A \times A$ on $A$ is called transitive, if we have
$$(a,b),(b,c) \in R \Rightarrow (a,c)$$
for all $a,b,c \in A$.
Your initial set is $R = \{(1,2),(2,3),(3,4),(4,1)\}$. The way you described your approach is basically the way to go. Just go through the set and if you find some $(a,b),(b,c)$ in it, add $(a,c)$.
The way the answer is given is a little bit confusing because it already tries to be explanatory :) The thing is, that they mean unions $\cup$ instead of compositions $\circ$. BUT they are writing it as a union to emphasize the steps taken in order to arrive at the solution:
- Step: Look at $R$. Since $(1,2),(2,3) \in R$, we need to add $(1,3)$ to the set. Analogously we have to add $(2,4),(3,1)$ and $(4,2)$. Lets call this new set $R_1$.
- Step: Now we have $R \subset R_1$, but $R_1$ is not yet transitive because $(1,3),(3,4) \in R_1$ but $(1,4)\notin R_1$. Hence, we have to add $(1,4)$ to $R_1$ and so on...
If we keep going we end up with the complete relation $R^+ = A \times A$ where $A = \{1,2,3,4\}$, i.e. $R^+$ contains ALL possible pairs of $1,2,3,4$.
By the way: I really like the idea to visualize the relation as a graph.
Best Answer
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.