 "shortcuts" for all currently possible directed walks.  That is to say, it is the smallest relation which is transitive and contains the original relation as a subset.

For your relation, it is a much smaller example.  Try drawing it out.  There are four vertices, $v_1,v_2,v_3,v_4$, and the corresponding graph will start with $5$ edges (one of which is a loop from $3$ to itself).  What shortcuts are missing for directed walks that we can perform?

 We can do a walk from $1$ to $3$ and from $3$ back to $1$ again, so we need a shortcut from $1$ to $1$ so we are missing $(1,1)$ in the transitive closure.  What else?

$~$

 We can't actually travel from $1$ to $2$.  Remember, we can't travel backwards along an edge... you should see that once you are at $1$ you can only stay at $1$ or go to $3$ or travel back and forth between $1$ and $3$.  So, we don't need a shortcut from $1$ to $2$ since there weren't any directed walks we could take from $1$ to $2$ to begin with.


As an additional aside, many of the usual properties we are interested in for relations can be described quite easily using the language of graph theory.


*

*Reflexivity:  All vertices have a loop.  (Originally: for all $x\in X$ we have $(x,x)\in R$)

*Symmetry:  All arrows (not loops) if any exist are double-sided.  That is, if you can go from $a$ to $b$ in a single step, then you can go from $b$ to $a$ in a single step.  (Originally: if $(a,b)\in R$ then $(b,a)\in R$).

*Antisymmetry: All arrows (not loops) if any exist are single-sided.  That is, if you can go from $a$ to $b$ (with $b$ different than $a$) in a single step, then you can not go from $b$ to $a$ in a single step.  (Originally: if $(a,b)\in R$ and $(b,a)\in R$ then $a=b$)

A: To augment JMoravitz's answer, we can also borrow a tool from graph theory known as adjacency matrices. Let's first represent the relation as a table:
$$\begin{array}{|c|c|}
\hline & \color{red}1 & \color{red}{2} & \color{red}{3} & \color{red}4 \\
\hline \color{red}1 \\
\hline \color{red}2 \\
\hline \color{red}3 \\
\hline \color{red}4 \\
\hline
\end{array}.$$
What I want to do is fill in the table with a $1$ if $\color{red}a$ is related to $\color{red}b$, in the $\color{red}a$ row and $\color{red}b$ column. Otherwise, I want to put a $0$ there. For example, $(\color{red}1, \color{red}3)$ is in your relation, so we'd put
$$\begin{array}{|c|c|}
\hline & \color{red}1 & \color{red}{2} & \color{red}{3} & \color{red}4 \\
\hline \color{red}1 & & & 1 \\
\hline \color{red}2 \\
\hline \color{red}3 \\
\hline \color{red}4 \\
\hline
\end{array}.$$
Filling out the rest of the table, we get
$$\begin{array}{|c|c|}
\hline & \color{red}1 & \color{red}{2} & \color{red}{3} & \color{red}4 \\
\hline \color{red}1 & 0 & 0 & 1 & 0 \\
\hline \color{red}2 & 0 & 0 & 0 & 1 \\
\hline \color{red}3 & 1 & 0 & 1 & 0 \\
\hline \color{red}4 & 1 & 0 & 0 & 0 \\
\hline
\end{array}.$$
Now, let's reject the leading row and column of the table to obtain the adjacency matrix:
$$A = \begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}.$$
Now, here's the cool bit: if we compute $A^2 + A$, we can easily find the transitive closure. Check it out:
\begin{align*}
A^2 + A &=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix} +
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix} \\ &=
\begin{pmatrix}
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & 2 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix} +
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix} \\ &=
\begin{pmatrix}
1 & 0 & 2 & 0 \\
1 & 0 & 0 & 1 \\
2 & 0 & 3 & 0 \\
1 & 0 & 1 & 0
\end{pmatrix}.
\end{align*}
What we do now is "flatten" the result, i.e. replace any integers greater than $0$ with $1$:
$$\begin{pmatrix}
1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 \\
1 & 0 & 1 & 0
\end{pmatrix}.$$
The result is the adjacency matrix for the transitive closure. For example, because there is a $1$ in the $\color{red}4$th row and $\color{red}3$rd column, it means $(4, 3)$ lies in the transitive closure. On the other hand, because there is a $0$ in the $\color{red}3$rd row and $\color{red}4$th column, $(3, 4)$ does not lie in the transitive closure.
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