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I've tried to prove it using the triangle inequality but can get nowhere. If you could supply a proof or counterexample I would be pleased.

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    $\begingroup$ No, it is not.${}$ $\endgroup$
    – user239203
    Aug 10, 2019 at 20:41
  • $\begingroup$ Sure, but R always has the natural ordering, which is what I was referring to when I said x>0 y<0 $\endgroup$
    – Aphyd
    Aug 11, 2019 at 4:29

3 Answers 3

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Here is a more natural counterexample.

Consider the continuous function $$f(x)=2|x|.$$

The graph of $f$, the set $V=\{(x,f(x)):x\in\mathbb R\}$, as a subspace of $\mathbb R^2$, is homeomorphic to $\mathbb R$; the mapping $x\mapsto(x,f(x))$ is a homeomorphism from $\mathbb R$ to $V$.

The Euclidean metric $d_2((x_1,y_1),((x_2,y_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ on $\mathbb R^2$, restricted to $V$, induces the natural topology of $V$ as a subspace of $\mathbb R^2$. Note that $(-1,2),(0,0),(1,2)\in V$, and $$d_2((1,2),(0,0))=\sqrt5\gt2=d_2((1,2),(-1,2)).$$ Transferring this metric from $V$ to $\mathbb R$, we see that $$d(x_1,x_2)=d_2((x_1,f(x_1)),(x_2,f(x_2)))=\sqrt{(x_1-x_2)^2+4(|x_1|-|x_2|)^2}$$ is a metric which induces the natural topology of $\mathbb R$, and $$d(1,0)=\sqrt5\gt2=d(1,-1).$$

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This cannot be true, because we can compose a metric with any permutation of $\mathbb R$ to obtain another metric. In particular, for all distinct $x, y, z \in \mathbb R$ we can find a permutation of $\mathbb R$ that sends $x$ to a positive number, $y$ to a negative number and $z$ to $0$.

The statement would thus imply that for every metric on $\mathbb R$ and all $x, y, z$ distinct, we have $d(x, z) \leq d(x, y)$. Changing the role of $y$ and $z$ implies $d(x, y) = d(x, z)$.

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  • $\begingroup$ That's unfortunate. Kind of breaks the intuition of what distance means. $\endgroup$
    – Aphyd
    Aug 10, 2019 at 21:10
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    $\begingroup$ @Aphyd Not really. An arbitrary metric on $\mathbb R$ just treats $\mathbb R$ as a set of points without any particular ordering, so you wouldn't expect $y < 0 < x$ to imply anything special. $\endgroup$
    – user169852
    Aug 10, 2019 at 21:13
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    $\begingroup$ @Aphyd It would have been a better question if, instead of "an arbitrary metric", you had asked about "any metric which induces the standard topology on $\mathbb R$." The answer would still be negative, but the counterexamples would be more natural and interesting. $\endgroup$
    – bof
    Aug 10, 2019 at 23:09
  • $\begingroup$ What about in a normed vector space? $\endgroup$
    – Aphyd
    Aug 11, 2019 at 2:28
  • $\begingroup$ Sorry I meant if we restrict our metrics on R to those induced by a norm is it true that d(x,0) is less than or equal to d(x,y) for x>0 y<0 $\endgroup$
    – Aphyd
    Aug 11, 2019 at 3:01
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Can't prove what isn't true.

Let $d(x,y) = 0$ if $x=y$. $d(x,y) = 1$ if $x,y$ are "on the same side of $0$" (that is if $0\le x$ and $0\le y$ or if $y \le 0; x\le 0$) and $d(x,y)=\frac 34$ if $x < 0 < y$ or $y< 0 < x$.

This is a metric: $d(x,y) = \{0,\frac 34, 1\} \ge 0$ and $d(x,y) = 0\iff x=y$ and $d(x,y) = d(y,x)$ and

$0 = d(x,y) \le d(x,z) + d(z,y)$ if $x=y$.

If $x,y$ are on the same "side of the zero" then $d(x,y)=1$ and $d(x,z)=d(z,y) = 1,\frac 34$ depending upon whether $z$ is or is not on the same side of the zero as $x$ and $y$. Either way $d(x,y) < \frac 34 + \frac 34 \le d(x,z) + d(z,y)$.

And if $x,y$ are on opposite sides of the zero, then if $z = 0$ we have $\frac 34 = d(x,y) < 1+1 = d(x,0) + d(0, y)$. If $z=x$ or $z=y$ with have $\frac 34 = d(x,y) = \frac 34 + 0\le d(x,z) + d(z,y)$. Otherwise $z$ is on the same side of one of $x$ or $y$ and the opposite sid of the other so $\frac 34 = d(x,y) < 1+\frac 34 = d(x,z) + d(z,y)$.

So triangle inequality holds.

So $d$ is a metric. And $1 = d(x,0) > d(x,y) = \frac 34$.

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