First of all, if this is how you define the transitive closure, then the proof is over. But you may still want to see that it is a transitive relation, and that it is contained in any other transitive relation extending $R$.
To the second question, the answer is simple, no the last union is not superfluous because it is infinite. Every step contains a bit more, but not necessarily all the needed information.
So let us see that $R^+$ is really transitive, contains $R$ and is contained in any other transitive relation extending $R$.
- Clearly $R\subseteq R^+$ because $R=R_0$.
- If $x,y,z$ are such that $x\mathrel{R^+} y$ and $y\mathrel{R^+}z$ then there is some $n$ such that $x\mathrel{R_n}y$ and $y\mathrel{R_n}z$, therefore in $R_{n+1}$ we add the pair $(x,z)$ and so $x\mathrel{R_{n+1}}z$ and therefore $x\mathrel{R^+}z$ as wanted.
- If $T$ is a transitive relation containing $R$, then one can show it contains $R_n$ for all $n$, and therefore their union $R^+$. To see that $R_n\subseteq T$ note that $R_0$ is such; and if $R_n\subseteq T$ and $(x,z)\in R_{n+1}$ then there is some $y$ such that $(x,y)\in R_n$ and $(y,z)\in R_n$. Since $R_n\subseteq T$ these pairs are in $T$, and since $T$ is transitive $(x,z)\in T$ as well.
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
Start by writing out $R$ explicitly:
$$R=\{\langle 0,1\rangle,\langle 1,4\rangle,\langle 2,7\rangle,\langle 3,10\rangle,\langle 4,13\rangle,\langle 5,16\rangle,\langle 6,19\rangle,\langle 7,22\rangle,\langle 8,16\rangle,\langle 8,25\rangle,\langle 9,28\rangle\}$$
Now look for the ‘linked’ pairs, like $\langle 0,\color{brown}1\rangle$ and $\langle\color{brown}1,4\rangle$: transitivity says that when you have linked pairs like that in the relation, you must also have corresponding the ‘shortcut’ pair, in this case $\langle 1,4\rangle$. Here the linked pairs are:
$$\begin{align*} &\langle 0,1\rangle\quad\text{and}\quad\langle 1,4\rangle\;,\\ &\langle 1,4\rangle\quad\text{and}\quad\langle 4,13\rangle\;,\text{ and}\\ &\langle 2,7\rangle\quad\text{and}\quad\langle 7,22\rangle\;, \end{align*}$$
so we have to add the shortcut pairs $\langle 0,4\rangle$, $\langle 1,13\rangle$, and $\langle 2,22\rangle$.
Now repeat the process: for example, we now have the linked pairs $\langle 0,4\rangle$ and $\langle 4,13\rangle$, so we need to add $\langle 0,13\rangle$. When you finish a second pass, repeat the process again, if necessary, and keep repeating it until you have no linked pairs without their corresponding shortcut.