3

Obviously it has to satisfy the following:

1) For all $x,y\in X$, $0\le d(x,y)$. (positivity)

2) For all $x,y\in X$, $d(x,y)=d(y,x)$. (symmetry)

3) For all $x,y,z\in X$, $d(x,y)\le d(x,z)+d(z,y)$. (triangle-inequality)

This is a homework problem and I'm not sure where to even start. I'm new to the concept of metric spaces and would appreciate any help/direction.

If x=y, then $\rho(x,y)=\rho(x,x)=(x-x)^2=0$. I'm assuming that will suffice for (1).

Michael Hardy
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mmh0015
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4 Answers4

4

$d(x,y)=(x-y)^2$

checking for 3rd axiom:

$$(x-y)^2+(y-z)^2 =x^2 -2xy + y^2 +y^2 -2yz +z^2\\=x^2 -2xz +z^2 +2xz -2xy + y^2 +y^2 -2yz\\=(x-z)^2+2x(z-y)-2y(z-y)\\=(x-z)^2+2(z-y)(x-y) $$ It means $$(x-z)^2=(x-y)^2+(y-z)^2-2(z-y)(x-y)$$ Now to satisfy the third axiom $2(z-y)(x-y)\ge0$ which is not possible say for $x<y<z$.

arghya
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1

Let's look at 1) as an example. You want to prove/disprove that $$\rho(x,y) = (x-y)^2 \geq 0$$ for any pair of real numbers $x$and $y$. Since the square of something is always non-negative, we can see that this holds.

But for $\rho$ to be a metric, 2) and 3) must hold as well! Try proving 2) by noting that $$(x-y)^2 = (-y + x)^2 = ((-1)(y - x))^2 = (-1)^2(y-x)^2 = (y-x)^2$$

For 3), try plugging in the specific values that David Mitra mentioned in the comments of your question. Do you see that this is a counter-example? What is your final conclusion about $\rho$?

EDIT: There is a fourth axiom which you forgot to mention: That $d(x,y) = 0$ happens if and only if $x=y$.

Ulrik
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  • Thank you for bringing to my attention the fourth axiom. After using specific values for 2) and 3), my conclusion is that they both hold and thus, $\rho$ is a metric. – mmh0015 Jan 13 '14 at 19:47
  • Although I'm confused about the counter-example. Are you able to explain it to me? – mmh0015 Jan 13 '14 at 19:48
  • So you checked that 2) and 3) were true for some specific values, and concluded that they were true for all values? – Ulrik Jan 13 '14 at 19:49
  • That is true, I would have to prove it holds for all values. – mmh0015 Jan 13 '14 at 19:59
  • Yes. Using the definition of your metric and the fact that $(x-y)^2 = (y-x)^2$, the proof of 2) just pops out. Can you see this? For 3), try plugging in $x = 0$, $y = 1$ and $z = 1/2$. Does the equation in 3) hold for these values? – Ulrik Jan 13 '14 at 20:02
  • Oh yes, the proof of 2) is clear. Also, when I test x=0, y=1, and z=1/2, 3) holds. I'm assuming that it is a counter-example and I'm doing something wrong. This is my work... – mmh0015 Jan 13 '14 at 22:16
  • $d(0,1) = 1$, $d(0,1/2) = 1/4$, $d(1/2,1) = 1/4$. Surely $1 \leq 1/4 + 1/4 = 1/2$ doesn't hold? – Ulrik Jan 13 '14 at 22:18
  • d(0, 1)<=d(0, 1/2)+d(1/2, 1) implies |0-1|<=|0-1/2|+|1/2-1| implies 1<=1/2 + 1/2 implies 1<=1. – mmh0015 Jan 13 '14 at 22:20
  • Oh I see, I need to use (x-y)^2 instead of absolute value – mmh0015 Jan 13 '14 at 22:21
  • Therefore 3) does not hold and thus, $\rho$ is not a metric. – mmh0015 Jan 13 '14 at 22:22
  • Yes, that's right! – Ulrik Jan 13 '14 at 22:23
  • Thank you so much for all your help! I now understand it much better. – mmh0015 Jan 13 '14 at 22:28
0

Let's check the triangle-inequality : let $x=0$, $y=1$, $z=2$, then $\rho(x,y)=1$, $\rho(y,z)=1$, and $\rho(x,z)=4$. From "triangle-inequality" we should have $$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$$ i.e. $$4 \leq 1 + 1$$ and that's not true, therefore $\rho$ is not a metric.

Arnaud D.
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Mahmood Al Sarraj
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0

$$ρ(x,z)=(x−z)^2=(x−y+y-z)^2=(x−y)^2 +(y−z)2 +2(z−y)(x−y)= ρ(x,y)+ρ(y,z)+2(z−y)(x−y),$$ $$ρ(x,z)= ρ(x,y)+ρ(y,z)+2(z−y)(x−y)$$

So triangular inequality is not satisfied in general.

sam wolfe
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ummfafool
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