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Thread: Discrete metric

  1. #1
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    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.
    Last edited by Connected; Mar 26th 2011 at 05:42 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Connected View Post
    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?
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  3. #3
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    No, I don't get it well.

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

    Thanks for your help.
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  4. #4
    Senior Member Tinyboss's Avatar
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    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?
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