# In $\mathbb{R}$ with an arbitrary metric, is it true that $d(x, 0) \leq d(x,y)$ for all $x > 0$, $y < 0$?

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.

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

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 $$d(x, y) = d(x, z)$$.

• That's unfortunate. Kind of breaks the intuition of what distance means. – Aphyd Aug 10 at 21:10
• @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. – Bungo Aug 10 at 21:13
• @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. – bof Aug 10 at 23:09
• What about in a normed vector space? – Aphyd Aug 11 at 2:28
• 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 – Aphyd Aug 11 at 3:01

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 $$d(1,0)=\sqrt5\gt2=d(1,-1).$$

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 $$1 = d(x,0) > d(x,y) = \frac 34$$.