Using the function

$f(x)= \begin{cases} \mathtt{true}, & \text{if x = 0}.\\\\ \mathtt{false}, & \text{if  x > 0}. \end{cases}$

we can define \$$\mathcal{X}(x, y) \$$ for our any \$$x, y \in \mathcal{X} \$$ in our "regions of the world" preorder:

\begin{align}
d(\texttt{Boston}, \texttt{Boston}) = 0,&\quad \mathcal{X}(\texttt{Boston}, \texttt{Boston}) = \mathtt{true} \\\\
d(\texttt{Boston}, \texttt{US}) = 0,&\quad \mathcal{X}(\texttt{Boston}, \texttt{US}) = \mathtt{true} \\\\
d(\texttt{Boston}, \texttt{Spain}) \neq 0,&\quad \mathcal{X}(\texttt{Boston}, \texttt{Spain}) = \mathtt{false} \\\\
d(\texttt{US}, \texttt{Boston}) \neq 0,&\quad \mathcal{X}(\texttt{US}, \texttt{Boston}) = \mathtt{false} \\\\
d(\texttt{US}, \texttt{US}) = 0,&\quad \mathcal{X}(\texttt{US}, \texttt{US}) = \mathtt{true} \\\\
d(\texttt{US}, \texttt{Spain}) \neq 0,&\quad \mathcal{X}(\texttt{US}, \texttt{Spain}) = \mathtt{false} \\\\
d(\texttt{Spain}, \texttt{Boston}) \neq 0,&\quad \mathcal{X}(\texttt{US}, \texttt{Boston}) = \mathtt{false} \\\\
d(\texttt{Spain}, \texttt{US}) \neq 0,&\quad \mathcal{X}(\texttt{US}, \texttt{US}) = \mathtt{false} \\\\
d(\texttt{Spain}, \texttt{Spain}) = 0,&\quad \mathcal{X}(\texttt{US}, \texttt{Spain}) = \mathtt{true} \\\\
\end{align}

From this we can create the Hasse diagram for the preorder:

![](https://image.ibb.co/nHfh0J/Screenshot_20180619_012121.png)