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Does $x \le y \le z$ imply $d(x, y) \le d(x, z)$?!

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  • $\begingroup$ It seems like the answer would be yes if $(X,d)$ were a certain kind of connected complete metric space called a length space people.math.ethz.ch/~lang/LengthSpaces.pdf. A length space is a metric space $(X,d)$ such that for all $x,y\in X,\epsilon>0$ there is some $z\in X$ where $d(x,z)<d(x,y)/2+\epsilon$ and $d(y,z)<d(x,y)/2+\epsilon.$ But then again, it seems like such a metric space would be isometric to some closed interval in the real numbers. $\endgroup$
    – Joseph Van Name
    Commented Aug 30, 2017 at 17:41

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No. Consider the set $X = \{x,y,z\}$ with the order $x < y < z$ and the metric $d(x,y) = 3$, $d(y,z) = 2$, and $d(x,z) = 1$.

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  • $\begingroup$ Thanks for both of you. What about infinite spaces, in particular uncountable ones, do separation axioms play rule here?! $\endgroup$
    – M. Alkadhi
    Commented Aug 30, 2017 at 13:17
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    $\begingroup$ You can take $Y = X \amalg \mathbb{R}_{\geq 0}$ with the usual order and metric on $\mathbb{R}_{\geq 0}$ and $x < a$ and $d(x,a) = a + 10$ for all $x \in X$ and $a \in \mathbb{R}_{\geq 0}$. Then both the order and the metric induce the same topology on $Y$: it is just the coproduct of $\mathbb{R}_{\geq 0}$ and the discrete space on $X$. This space is uncountable and all metrizable spaces are $T_6$ spaces. $\endgroup$
    – Valery Isaev
    Commented Aug 31, 2017 at 7:40
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Consider the unit interval $[0,1]$ with its usual ordering, embedded into $\mathbb{R}^2$ as follows: enter image description here

Let $d$ be the induced metric from $\mathbb{R}^2$. This metric is compatible with the order topology, yet clearly $0 \le 1/2 \le 1$ while $d(0,1/2) \gg d(0,1)$.

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  • $\begingroup$ I didn't understand what do you mean by "compatable with the order topology" yet it doesn't preserve order !!!. $\endgroup$
    – M. Alkadhi
    Commented Aug 30, 2017 at 22:09
  • $\begingroup$ I sould emphasize that we only consider the metric which generate the order topology under consideration. $\endgroup$
    – M. Alkadhi
    Commented Aug 30, 2017 at 22:14
  • $\begingroup$ @M.Alkadhi: The induced metric does generate the order topology. That is what I mean by "compatible". I do not know what you mean by "preserve order". $\endgroup$
    – Nate Eldredge
    Commented Aug 30, 2017 at 22:58

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