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I have 2 distance functions $d(x,y)=|x^2-y^2|$ and $d(x,y)=|x^3-y^3|$ and I am trying to prove that they are metrics on $\mathbb R$, or give a counterexample that they are not metrics on $\mathbb R$.

I managed to prove the first 3 properties for the 2nd function namely:

  1. $d(x,y)\ge0$

  2. $d(x,y)=0$ iff $x=y$

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

and I now have shown that the 1st function is not a metric.

But for the triangle inequality property, $d(x,y)\le d(x,z) + d(z,y)$, Im stuck on it. How would go about proving it for $d(x,y)=|x^3-y^3|$? Thanks.

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    you made a mistake on the second property for the first of the mentioned maps, as it is not satisfied by that one.2012-02-13
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    regarding property 2: what is $d(1,-1)$ for your first function?2012-02-13
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    Thank you both, I think when I first tackled the question, I didnt think of it with negative numbers, and I was just more focused on the fourth property. Thanks again. May I ask, how would I prove the triangle inequality property holds for the 2nd function?2012-02-13
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    Hint: In the inequality you have to prove, try substituting x^3 = a, y^3 = b, z^3 = c. See anything familiar?2012-02-13
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    Thanks, is this correct? $d(x,y)=|x^3-y^3| = |a-b| \le |a-c| + |c-b| \le d(x,z) + d(z,y)$?2012-02-13
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    See also: http://math.stackexchange.com/questions/965818/which-properties-must-a-function-f-fulfill-for-d-to-be-a-metric2015-12-07

2 Answers 2

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I'll write up an answer, so that it is not on the Unanswered Category:

A function $d: X \times X \to \mathbb R$ is said to be a metric on $X$ if:

  1. (Non-negativity) $d(x,y) \ge 0$ for all $x,y \in X$
  2. (Definiteness) $d(x,y) =0 \iff x=y$
  3. (Symmetry) $d(x,y)=d(y,x)$ for all $x,y \in X$
  4. (Triangle Inequality) $d(x,y) \le d(x,z)+d(z,y)$ for all points $x, y, z \in X$

Consider the following functions $d$ as metrics on $\mathbb R$.

  • $d(x,y)=|x^2-y^2|$

It is NOT a metric. Note that "definiteness axiom" fails here: $d(1,-1)=0$ but obviously, $1 \neq -1$

  • $(d(x,y)= |x^3-y^3|$

It is a metric. Note that the triangle inequality follows from that on the Euclidean Metric on $\mathbb R$.

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    Thanks, how would I show the triangle inequality property? Is this correct? $d(x,y)=|x^3-y^3| = |a-b| \le |a-c| + |c-b| \le d(x,z) + d(z,y)$?2012-02-13
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    Yes, exactly! :)2012-02-13
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    @William: Except your last $\leq$ would be less confusing as an $=$. You are just using the triangle inequality for the absolute value: $d(x,y) = |x^3-y^3| = |(x^3-z^3)+(z^3-y^3)| \leq |x^3-z^3|+|z^3-y^3| = d(x,z)+d(z,y)$.2012-02-13
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    My eye missed that @Arturo. Thank You. A reminder about that $G/Z(G)$. That ' bounded below' needs to be fixed, I think. Please let me know if I am wrong or otherwise :)2012-02-13
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    @KannappanSampath: Really? As far as I can tell, it was fixed seven hours ago (three hours after your comment here).2012-02-14
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    @ArturoMagidin I had realised that! I am sorry. Strangely, I got no ping. So, I had gone there to see and it was all there :) Sorry once again!2012-02-14
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You can think about it in two ways, both following from Lopsy's hint. I'm not writing any new mathematics here, it's just a longer comment.

You can note that the function $f(x) = x^3$ is a homeomorphism of $\mathbb{R}$. Actually all you need is that it is injective, but using the fact that it is homeo one can show that the new metric leads to the same topology as the standard metric. Anyway - if one has a metric $d_Y$ on $Y$ and and injective function $g:X \to Y$ then the function $d_X:X\times X \to \mathbb{R}$ defined by $d_X(x,y) = d_Y(g(x), g(y))$ is a metric on $X$. It is an easy excercise and you don't have to worry about any special properties of the cubic function. In our case: $d_\mathbb{R}(x,y) = |x^3 - y^3|$ is taking the metric from $Y=f(\mathbb{R})$ to $X=\mathbb{R}$ with function $g=f$: $$d_\mathbb{R}(x,y) = d_{f(\mathbb{R})}(f(x), f(y)),$$ where $d_{f(\mathbb{R})}(a,b) = |a-b|$.

The second way - identical from the mathematical point of view but maybe useful for imagination - is looking at $d(x,y)$ as a metric on the graph of function $f$ defined as above. One projects the graph on the image and takes the metric from the image. It is important that the projection does not glue any points (because $f$ is injective).

Using this language - the first function $d(x,y)=|x^2-y^2|$ is not a metric because $f$ is not injective and glues $a$ and $-a$. When you thought about positive halfline then it was injective and everything was OK.

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    Very nice Answer. +12012-02-13
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    Thanks, very nice way to see it!2012-02-13