# Discrete metric

• Mar 26th 2011, 05:31 PM
Connected
Discrete metric
Let $\displaystyle (E,d)$ be the discrete space.

Compute $\displaystyle S(a,r)$ for $\displaystyle r>1,$ and $\displaystyle S(a,r)$ for $\displaystyle r\le1.$

I know that $\displaystyle d(x,y)$ is defined to be $\displaystyle 1$ for $\displaystyle x\ne y$ and $\displaystyle 0$ for $\displaystyle x=y,$ but I don't know exactly how to use that to solve the problem.
• Mar 26th 2011, 05:40 PM
Drexel28
Quote:

Originally Posted by Connected
Let $\displaystyle (E,d)$ be the discrete space.

Compute $\displaystyle S(a,r)$ for $\displaystyle r\ge1,$ and $\displaystyle S(a,r)$ for $\displaystyle r\le1.$

I know that $\displaystyle d(x,y)$ is defined to be $\displaystyle 1$ for $\displaystyle x\ne y$ and $\displaystyle 0$ for $\displaystyle x=y,$ but I don't know exactly how to use that to solve the problem.

I assume that $\displaystyle S(a,r)$ is the 'sphere' (that is an uncommon noation) of radius $\displaystyle r$ centered at $\displaystyle a$. I think you're overthinking your problem. $\displaystyle S(a,r)$ has a simple formulation. Think about fixing this one point $\displaystyle a$ then one can think (purely heuristically) as the situation being analgous to $\displaystyle a$ being the origin in $\displaystyle \mathbb{R}^2$ and $\displaystyle E-\{a\}$ being the unit circle. With this in mind, is the solution clear?
• Mar 26th 2011, 05:43 PM
Connected
No, I don't get it well.

I understand a bit your reasoning, but is there a way to make it analytically?

If $\displaystyle S(a,r)=\{x\in E\mid d(x,a)<r\}$, then what happens in the two cases $\displaystyle r<1$ and $\displaystyle r\ge1$? In particular, what do you know if d(x,y)<1 in this metric?