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The following is a question of Metric Spaces by O'Searcoid (pg 19)

Suppose $X$ is a set and $d:X×X→\mathbb{R}$. Show that $d$ is a metric on $X$ if, and only if, for all $a,b,z ∈ X$, the two conditions $d(a,b) = 0 ⇔ a = b$ and $d(a, b) ≤ d(z, a) + d(z, b)$ are both satisfied.

The the first direction seems easy to demonstrate.

If $d$ is a metric, the first condition holds by definition, and by applying symmetric, the second condition follows from the triangle inequality.

The converse statement, however, is presenting some difficulties.

From what I can see, I must now demonstrate the the two conditions imply that $d$ is positive for all $ a \ne b$ and that $d$ is symmetric.

Suppose $d$ is not symmetric. Then this contradicts the first statements, since $d(a,b) \ne d(b,a)$ and if $a=b$, than $d(a,b) = 0$ but $d(b,a) \ne 0$. This also implies the triangle inequality by applying symmetry on the second condition in the question.

What I cannot figure out is how to demonstrate $d(a,b) \ge 0$. If we allow $d$ to be negative, I do not see how this may lead to a contradiction of these statements.

One thing I thought of was, suppose $d(a,b) = 0$. Then this implies that $0 ≤ d(z, a) + d(z, b)$ and so these distance functions with $z$ must be positive. We could form any "intermediary" distance function we please regardless of the left-hand-side distance function. Thus in order to solve the triangle inequalities for all $a$ given $d(a,a)$ we require each $d$ to be positive.

Is my reasoning valid?

GovEcon
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    Perhaps I'm missing something obvious, but in the converse statement, the proof you've given showing that $d$ is symmetric doesn't seem right. If $d$ is not symmetric, then that implies that $\exists a, b$ such that $d(a, b) \neq d(b, a)$, not that $d(a, b) \neq d(b, a) \forall a, b$. I may be misunderstanding you or missing something myself, though. – Alex Wertheim Jul 03 '13 at 02:57
  • @AWertheim I believe you are correct. I am uncertain about both parts of my proof. I just was unable to determine what I was uncertain about. – GovEcon Jul 03 '13 at 02:59

2 Answers2

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We have $d(a, b) ≤ d(z, a) + d(z, b)$ so:

  • If $a=b$ we have $0\leq 2d(z,a)$ and then $d(z,a)$ is non negative for all $z$ and $a$
  • Now replace $z$ by $b$ we find $d(a,b)\leq d(b,a)$ and switching $a$ and $b$ gives $d(a,b)\geq d(b,a)$ so we find $d(a,b)=d(b,a)$.
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Starting with the given inequality $d(a,b)\leq d(z,a)+d(z,b)$, specialize to $z=b$ to get symmetry, and therefore the usual triangle inequality. Then go back to general $z$ but specialize to $a=b$ (and use symmetry) to get nonnegativity.

Andreas Blass
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