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Given the discrete metric. I have to show the different environments (amount of shape $\{x ∈ X: d(x,x_0) < r\}$) for different $x_0$ and $r$. My thoughts: $r$ must equal to $1$ because we have the discrete metric, so the environment must look like $E(x, x_0) < 1$. Now whats happening with the "different" $x_0$?

Or Shahar
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ggg7
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I hope I'm understanding you right, but pay attention to the $\textbf{name}$ of this metric, you are talking abut the discrete metric, which is corresponding with the discrete topology (the open sets are all the subsets of $X$).

This metric is not the euclidean metric, it is pretty hard to visualise this metric, but it is possible to visualise the open balls:

Let $B(x,r)$ be the open ball with radius $r$ at $x\in X$. $$B(x,r):=\{y\in X\mid d(x,y)<r\}$$

Hence: $\forall x\in X:$

If $0\leq r<1$, $B(x,r)=\{x\}$. Why? Because $B(x,r):=\{y\in X\mid d(x,y)<r\}$ and $0\leq r<1$, but the metric is discrete, hence $\forall x\neq y\in X: d(y,x)=1$, but $0\leq r<1$!, which means $B(x,r)=\{y\in X\mid d(x,y)<r<1\}=\{x\}$

If $1\leq r$, $B(x,r)=X$. Why? Because $B(x,r):=\{y\in X\mid d(x,y)<r\}$ and $r\geq1$, but the metric is discrete, hence $\forall x\neq y\in X: d(y,x)=1$, and in particular $d(y,x)\geq 1$

Or Shahar
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  • There is a useful case that can be visualised: the vertices of a suitably sized equilateral triangle or more generally the vertices of a (suitably scaled) standard $n$-simplex inherit the discrete metric from the euclidean metric. – Rob Arthan Mar 14 '21 at 16:30
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    thanks! very understandable! – ggg7 Mar 15 '21 at 09:13
  • @jz97x, I'm glad to hear that I answerd your question, please accept my answer by pressing the V button if you think it was helpful – Or Shahar Mar 15 '21 at 10:00
  • Did you mean 0 < r ≤ 1 and 1 < r when you have compared the two cases for r? I have just seen a similar Proof but with 0 < r ≤ 1 and 1 < r and the same result as yours – ggg7 Mar 15 '21 at 18:02
  • @jz97x No, as I wrote, the case when the radius is less than 1 the open ball is trivial, but when the radius is equal or bigger than 1 the open ball is the whole space. – Or Shahar Mar 15 '21 at 22:25