Metric Space Axioms

Question

I've been reading about metric spaces recently and had a question about the definition. Everywhere I've seen, we take a metric space to be a set $E$ equipped with a metric, a binary map $d:E\times E\mapsto \mathbb{R}$, where $\forall x, y, z\in\mathbb{R}$, $d$ satisfies:

$d(x, y) \geq 0$
$d(x, y) = 0 \iff x = y$
$d(x, y) = d(y, x)$
$d(x, z) \leq d(x, y) + d(y, z)$

It isn't difficult to see that axioms $2$, $3$, and $4$ imply $1$, so $1$ doesn't add anything to the definition and is superfluous. Why then do we keep 1 as an axiom in the definition of a metric instead of it being a corollary to the definition?

or better yet, just define $d$ to be a function $d\colon E\times E \to \mathbb{R}_{\geq 0}$. — user259242, Nov 05 '16 at 03:56

score 4 · Answer 1 · answered Nov 05 '16 at 04:04

4

In fact, it is enough to require only that $$d(x,y)=0 \textrm{ if and only if } x=y$$ and $$d(y,z)\leq d(x,y) +d(x,z)$$

All other properties of the metric follow from these, including nonnegativity.

answered Nov 05 '16 at 04:04

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How does (3) follow from these? – Jonathan Hebert Jan 03 '20 at 20:47
Nevermind, I see the slight of hand now. ;) – Jonathan Hebert Jan 03 '20 at 20:53

score 1 · Answer 2 · answered Nov 05 '16 at 04:03

1

There is obviously no need ;

But since "Metric Space" deals with distance function and the distance between any two objects is always non-negative so the axiom is added to mark its significance.

answered Nov 05 '16 at 04:03

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Metric Space Axioms

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