Picturing the discrete metric? A chapter on topology talks about this and asks me to "think of a model for the discrete metric on $M$" where $M$ may have $1, 2, 3, 4$ points. Here is what I think:  
$2$ points = $2$ distinct points  
$3$ points = equilateral triangle  
$4$ points = tetrahedron
$1$ point = ??? A point by itself? But that would mean $1 = 0$...
The book goes on to say "imagine the discrete metric on $\mathbb{R}$", which I'm unable to grasp. Thanks for your help!
 A: I am not sure, why do you think about objects from continuous geometry, the following example shall work I guess. A finite graph is a finite set $V$ of points, and $E\subseteq V\times V$ are edges of this graph. We can endow $V$ with a metric by
$$
  d(x,y) = n\in \Bbb N_0\quad \Leftrightarrow \quad \text{any path from }x\text{ to }y\text{ passes at least }n\text{ edges}
$$
The graph is called fully connected if $E = V\times V$. Then $d$ brings a discrete topology on a fully connected graph: from any point on such graph, you can reach any other point passing by only $1$ edge. If there is only one point, of course you don't have to move anywhere.
A: Hint: Stop imagining your topologies as belonging to Euclidean space. If you were going to imagine a discrete metric on $\mathbb{R}$ by visualizing the placement in $\mathbb{R}^3 $, there simply wouldn't be any room!
In the case of $ \mathbb{R} $, imposing the discrete metric just results in a case where every point $ x \neq y $ is of distance 1 away from each other. It just refers to the points being "separated", you're not meant to visualize the points separated in Euclidean space.
