I have the following statement from my notes:
"Let $(X,d)$ be the discrete space i.e. any non-empty set with the discrete metric ($d_d(x,y)=1$ for all $x\neq y$). Then, amazingly, $B_1(x)=\{x\}$, a set consisting only of $x$, and $\tilde B_1(x)=X.$"
With the discrete space $(X,d)$, being the set $X$ with the discrete metric defined on it as,
$d_d(x,y)=0$ if $x=y$
and,
$d_d(x,y)=1$ if $x\neq y$
We also have $B_1(x)$, the open ball centered at $x$ with radius $1$, and $\tilde B_1(x)$ the closed ball centered at $x$ of radius $1$.
This is what I have so far to verify the first part of the statement (note, I am not asking how to show/prove this statement, so please bear with me):
Our open ball, $B_1(x)$, can be written equivalently as,
$$B_1(x)=\{y\in X:d(x,y)\lt1\}$$
And we know that the boundary of our ball will be all $y$ satisfying,
$$d(x,y)=1$$
So then $y$, intuitively, is on the boundary of the sphere, (and hence, not in the sphere) and $x\neq y$. This leaves us with the case of $x=x$. From the properties of a metric space, we know that,
$$d(x,x)=0$$
And of course,
$$0<1\implies x\in B_1(x)$$
Now here comes my first question; how is it that an open ball of radius $1$ can have only one element? I am finding it hard to conceptualize that this is the only element in the set. Can anybody help me to visualize this better and understand it more clearly?
Could somebody also help me to clarify, in a similar fashion, the second part of the statement regarding the closed ball, namely that $\tilde B_1(x)=X$?
Is the point that I am finding hard to grasp as simple as supposing that $x$ and $y$ are the only elements in $X$?
For an arbitrary element $x \in X$, what $B_1(x)={x}$ means is that the only element that is within a distance of $1$ of $x$ is $x$ itself, and what $\tilde B_1(x)=X$ means is that all elements in $X$ are within $1$ of $x$, or, since $x$ was arbitrary, that all elements of $X$ are within $1$ of each other. It all follows from the definition of the discrete metric.
– coldnumber Jul 25 '15 at 19:58