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.
I suspect that "Using the connectivity relation" to find the transitive closure is the following algorithm. Start with $T=\emptyset$ as a beggining of the transitive closure
- Choose a letter (say, start with a)
- See to which elements a are related in as first part, in this case we only have $(a,e)$ so $a$ is only related to $e$. Add all of these pairs to $T$
- For each new element found in the previous step see which pairs exist with these elements in the first part. Add $a$ together with the second element in these pairs to $T$. In our case $e$ is related to $a$, $b, c, e$ hence we add $(a,a), (a,b), (a,c) $ and $(a,e)$ to T.
- Repeat from step 3. for each new element found previously.
- Repeat from step 1 for each element.
Note that after we have done up to step 5. for $a$ T will consist of $\{(a,a),(a,b),(a,c),(a,d),(a,e)\}$
This method is essentially to write the graph which R forms, and see which elements you may reach from each node, and add these edges to the graph, why you may call it the connectivity relation.
Best Answer
Given any relation $R$ on $A$, the transitive closure may as well be described as the set $T$ consisting of all pairs $(a,b)\in A\times A$ such that there exists a finite sequence of elements $$ a=a_0, a_1, a_2, \dots, a_{k-1}, a_k = b \in A $$ with $$ (a_0,a_1), (a_1,a_2), (a_2,a_3), \dots, (a_{k-1}, a_k) \in R. $$ Note that when such a sequence exists for a given pair $(a,b)$, it follows that there also exists such a sequence without repetitions, since you can just remove the subsequence between the repeating elements.
Hence, you may describe the transitive closure $T$ as the set of all pairs $(a,b)$ such that there exists a non-repeating sequence in $A$ with the above property.
Now if $|A|=n$, the maximal length of a non-repeating sequence in $A$ is $n$ and hence for the transitive closure it is enough to include only up to $(n-1)$-fold compositions of elements in $R$.