Identical metric spaces

Question

My class was deleted, due to the virus.

I am supposed to solve the following problem:

Let $A$ be any set of positive real numbers. Construct a metric space $(X, d)$ so that the set $A$ is identical to the set of all distances of different points of the space $X$.

My solution: \begin{align*} X&=\left \{ 0 \right \}\cup A\\ d(x,y)&= \begin{cases} 0 & \text{if $x=y$,}\\ \max\{x,y\} & \text{if $x\neq y$.} \end{cases} \end{align*}

Is that correct? How should I continue?

You need to check that $d(x,y)=0$ if and only if $x=y$, that $d(x,y)=d(y,x)$ and the triangle inequality $d(x,y)+d(y,z) \ge d(x,z)$ has to be satisfied for all $x,y,z\in X$. — Christoph, Mar 18 '20 at 12:02
@Christoph for the triangle inequality I tried: x=z, x=y and y=z, is that enough? — Shelley, Mar 18 '20 at 12:10
Once you fix the typo and write $d(x,y)$ instead of $d(x,x)$, it seems right, but you need to check the metric space axioms. — Ian, Mar 18 '20 at 12:14

Kavi Rama Murthy · Accepted Answer · 2020-03-18T12:33:53.613

0

Triangle inequality: If $x,y,z$ are all distinct the n $x \leq \max \{x,z\}\leq d(x,z)+d(z,y)$ and $y \leq \max \{y,z\}\leq d(x,z)+d(z,y)$. Hence $d(x,y) =\max \{x,y\} \leq d(x,z)+d(z,y)$. The case when the points are not distinct is trivial.

edited Mar 18 '20 at 12:33

answered Mar 18 '20 at 12:02

Kavi Rama Murthy

311,013

1

Thanks, that is enough? – Shelley Mar 18 '20 at 12:17
Note that $d(x,y)$ is not always equal to $\max{x,y}$. – Christoph Mar 18 '20 at 12:31
@Christoph Triangle inequality trivially holds when the points are not distinct. – Kavi Rama Murthy Mar 18 '20 at 12:35

Identical metric spaces

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