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First, I am aware of how to show the following, my question concerns the reasoning of the standard metric definition.

Traditionally, the axioms defining a metric $\rho:X\times X \rightarrow \mathbb{R}$ where $X$ is a nonempty set, are given as:

  1. $\forall x, y\in X, \rho(x, y) = 0 \iff x = y$,

  2. $\forall x, y\in X, \rho(x, y) = \rho(y, x)$,

  3. $\forall x, y, z\in X, \rho(x, y)\leq \rho(z, x) + \rho(z, y)$,

  4. $\forall x, y\in X, 0 \leq \rho(x, y)$.

It is easy to see that axiom 4 follows from axioms 1 and 3. It is also fairly trivial to show that axiom 2 follows from axioms 1 and 3. It then seems logical to only require the two axioms

  1. $\forall x, y\in X, \rho(x, y) = 0 \iff x = y$,

  2. $\forall x, y, z\in X, \rho(x, y)\leq \rho(z, x) + \rho(z, y)$,

as a definition for a metric. Is there a good reason that this is not the case, besides historical continuity?

Edit: Thanks to Paul Frost for mentioning my original definition was for pseudometrics... The question has been updated.

Edit 2: As commenters have asked for a proof:

Axioms 1, 3 imply axiom 2: First, consider the triangle inequality for general $x, y, z\in X$. Then, $$\rho(x, y) \leq \rho(z, x) + \rho(z, y),$$ next, set $z = y$ to obtain $$\rho(x, y) \leq \rho(y, x) + \rho(y, y)\Rightarrow \rho(x, y) \leq \rho(y, x)$$ by axiom 1. Similarly, $$\rho(y, x) \leq \rho(z, y) + \rho(z, x),$$ and setting $z=x$ yields $$\rho(y, x) \leq \rho(x, y) + \rho(x, x)\Rightarrow \rho(y, x) \leq \rho(x, y)$$ so $\rho(x, y) = \rho(y, x)$.

We may then show Axioms 1, 2, and 3 imply 4. See this answer. It may be somewhat incorrect to say axioms 1 and 3 imply 4, but I don't see this as a problem, as 1 and 3 imply 2, and 1, 2, 3 imply 4. If I'm incorrect in that assumption, please let me know.

hoverless
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    You consider pseudometrics because you do not require $\rho(x,y) = 0 \Rightarrow x = y$. – Paul Frost Dec 23 '18 at 18:33
  • You're right, let me edit that requirement. My mistake! – hoverless Dec 23 '18 at 18:36
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    You say that it is "easy" to see that axiom 4 follows from axioms 1 and 3. Perhaps you would like to share your proof. – Xander Henderson Dec 23 '18 at 18:39
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    Your statement Is not quite the usual triangle inequality, which is $\forall x, y, z\in X.\rho(x, y) \le \rho(x, z) + \rho(z, y)$. Whether that affects your proof that symmetry follows from other axioms is unclear, because you haven't given that proof. Please give the proof. – Rob Arthan Dec 23 '18 at 18:41
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    Your way of stating the triangle inequality is not standard. The standard definition maintains the "alphabetical order" of $x,y,z$: $\forall x, y, z\in X, \rho(x, z)\leq \rho(x,y) + \rho(y,z)$. This standard version is used for interesting examples of "asymmetric metrics". – Lee Mosher Dec 23 '18 at 18:45
  • The "traditional" version of the triangle inequality conveys quite naturally the idea that "the distance in the path going directly from A to B is no more than the distance in any path going from A to B through a third point." – Mauro ALLEGRANZA Dec 23 '18 at 18:46
  • @XanderHenderson I have added a proof of axioms 1, 3 imply 2, and a link to a previous question where 1, 2, and 3 are used to prove 4. – hoverless Dec 23 '18 at 18:55
  • @LeeMosher So, is it more so the use of a nonstandard triangle inequality that is the issue here? I haven't run into such asymmetric metrics so I am unaware of how my definition would clash. – hoverless Dec 23 '18 at 18:55
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    There are even very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) \mapsto xy^{-1}$) but this will generalyl make the axiomatisation harder to understand and harder to generalise. – Rob Arthan Dec 23 '18 at 18:56
  • I think the last comment of @RobArthan is the real answer to this question. – Lee Mosher Dec 23 '18 at 18:57
  • @Lee Mosher: given your encouragement, I've upgraded my comment to an answer. I was a little concerned that the answer has to be a matter of opinion. – Rob Arthan Dec 23 '18 at 19:00
  • But it is more than value judgement, it's mathematical judgement. I was thinking about the same example of $xy^{-1}$, but you made an excellent additional point about the mathematical effect of these kinds of clever changes. – Lee Mosher Dec 23 '18 at 19:04
  • You don't seem to include $d(x,x)=0$ ... what stops us from having $d(x,y)=1$ for all $x,y$? (That is, in $(1)$ your "$\implies$" should be an "$\iff$.") – Noah Schweber Dec 23 '18 at 19:12
  • @NoahSchweber Whoops. Edited. – hoverless Dec 23 '18 at 21:49

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Your statement of the triangle inequality is not quite the same as the usual one, which says $d(x, y) \le d(x, z) + d(z, y)$. The usual definition supports the intuition that $d(x, y)$ measures the time (or cost or work) of getting from $x$ to $y$ with no assumption that getting from $x$ to $y$ involves the same amount of time (or ...) as the return journey.

There are very simple examples of asymmetric metrics: e.g., what you might call the morning rush-hour metric between a city $A$ and a suburb $B$ with $d(A, B) < d(B, A)$ and $d(A, A) = d(B, B) = 0$. See Lee Mosher's comment for many more interesting examples.

The answer to your question necessarily involves a matter of mathematical opinion. The usual axioms for metric spaces clearly separate several concerns. As often happens with axiomatisations, you can find an axiomatisation with fewer axioms or operators by making some clever changes (e.g., axiomatising a group in terms of the operation $(x, y) \mapsto xy^{-1}$) but this will generally make the axiomatisation harder to understand and harder to generalise.

Rob Arthan
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