Are the positive-semidefinite requirements necessary for metric space axioms?

[Bipolarity]

Quick question about the metric space axioms, is the requirement that the distance function be positive-semidefinite an axiom for metric spaces?

It seems that it can be proved from the other axioms (symmetry, identity of indiscernibles and the triangle inequality).

BiP

 

[Jorriss]

Assuming by 'identity of indiscernible' you mean d(x,y) = 0 iif x=y then yes, requiring that d(x,y) be nonnegative is redundant.

 

[SteveL27]

Quick question about the metric space axioms, is the requirement that the distance function be positive-semidefinite an axiom for metric spaces?

It seems that it can be proved from the other axioms (symmetry, identity of indiscernibles and the triangle inequality).

BiP

According to my understanding, the phrase "identity of indiscernibles" refers to the philosophical principle that two objects are the same if they share all properties. As Wiki puts it ...

The identity of indiscernibles is an ontological principle which states that there cannot be separate objects or entities that have all their properties in common. That is, entities x and y are identical if every predicate possessed by x is also possessed by y and vice versa.

http://en.wikipedia.org/wiki/Identity_of_indiscernibles

I've never heard it used mathematically; and in fact it is false about the real numbers. There are only countably many predicates (finite length strings over a finite or even countable alphabet); but there are uncountably many reals. It is the case mathematically that there are more real numbers than predicates. It follows that there are pairs of distinct real numbers that nevertheless agree on all predicates.

I hope I didn't totally miss your point here, but I've just never heard anyone use identity of indiscernibles in the context of metric spaces.

 

[R136a1]

Yeah, "identity of indiscrenables" is not a common phrase in mathematics.
And describing the metric as "positive semidefinite" is also not common. The terminology is used primarily for matrices or bilinear forms, but not for a metric.

Aside from this terminology issue, jorriss already gave the right answer.

 

[gufiguer]

Let X be a set. A function [; d \ : \ X \times X \to \mathbb{R} ;] is a metric for X if, and only if,
(i) d(x,x)=0, for all x in X
(ii) d(x,y)=d(y,x), for all x and y in X
(iii) [; (\forall x,y \in X) [x \neq y \to d(x,y) > 0] ;]
(iv) [; (\forall x,y,z \in X) [d(x,z) \leq d(x,y) + d(y,z)] ;]

And (i)-(iv) are independent axioms. You can prove this by constructing some simple models that is true three of them but fails in one.

 

[1MileCrash]

Let X be a set. A function [; d \ : \ X \times X \to \mathbb{R} ;] is a metric for X if, and only if,
(i) d(x,x)=0, for all x in X
(ii) d(x,y)=d(y,x), for all x and y in X
(iii) [; (\forall x,y \in X) [x \neq y \to d(x,y) > 0] ;]
(iv) [; (\forall x,y,z \in X) [d(x,z) \leq d(x,y) + d(y,z)] ;]

And (i)-(iv) are independent axioms. You can prove this by constructing some simple models that is true three of them but fails in one.

By (iv)
d(x,y) + d(y,x) >= d(x,x)

By (ii), is just
d(x,y) + d(x,y) >= d(x,x)

2d(x,y) >= d(x,x)

And so by (i)
2d(x,y) >= 0

d(x,y) >= 0

Which is (iii). You cannot construct a 'metric candidate' where (i), (ii), and (iv) hold but (iii) does not; (iii) is redundant, just like the OP said.

 

[R136a1]

By (iv)
d(x,y) + d(y,x) >= d(x,x)

By (ii), is just
d(x,y) + d(x,y) >= d(x,x)

2d(x,y) >= d(x,x)

And so by (i)
2d(x,y) >= 0

d(x,y) >= 0

Which is (iii). You cannot construct a 'metric candidate' where (i), (ii), and (iv) hold but (iii) does not; (iii) is redundant, just like the OP said.

Not exactly, notice that (iii) has a strict inequality. You just showed $\geq$. So the axioms gufiguer uses are different than what we usually use.

 

[1MileCrash]

Not exactly, notice that (iii) has a strict inequality. You just showed $\geq$. So the axioms gufiguer uses are different than what we usually use.

But it's the same, all I have to do is declare that x and y are not equal and the inequality does become strict, just like gufiguer's (iii).

 

[R136a1]

But it's the same, all I have to do is declare that x and y are not equal and the inequality does become strict, just like gufiguer's (iii).

You need an axiom to ensure that the inequality is strict. For example, if we set $d(x,y)=0$ for all $x$ and $y$, then this satisfies gufiguer's axioms except (iii) and it is positive.

 

[1MileCrash]

You need an axiom to ensure that the inequality is strict. For example, if we set $d(x,y)=0$ for all $x$ and $y$, then this satisfies gufiguer's axioms except (iii) and it is positive.

OK, now I see, gufiguer's first axiom is usually given as d(x,y) = 0 iff x=y. They are pretty different than the ones I've seen. I don't think they're correct, actually, d(x,y) = 0 for all x and y shouldn't really qualify as a metric..

When I think about the metric d(x,y) = 0 for all x and y, my only conclusion is that it generates the indiscrete topology. Consider any open ball in this metric space (X, G); it contains everything, it's just X. A nonempty, proper subset U of X cannot possibly be open because it cannot contain any open ball. So this metric generates the indiscrete topology.

But, my issue with this is that I know I proved this semester that the indiscrete topology is not metrizable. So how can gufiguer's axioms be correct, and still allow for this type of "metric" to exist?

 

[R136a1]

OK, now I see, gufiguer's first axiom is usually given as d(x,y) = 0 iff x=y. They are pretty different than the ones I've seen. I don't think they're correct, actually, d(x,y) = 0 for all x and y shouldn't really qualify as a metric..

When I think about the metric d(x,y) = 0 for all x and y, my only conclusion is that it generates the indiscrete topology. Consider any open ball in this metric space (X, G); it contains everything, it's just X. A nonempty, proper subset U of X cannot possibly be open because it cannot contain any open ball. So this metric generates the indiscrete topology.

But, my issue with this is that I know I proved this semester that the indiscrete topology is not metrizable. So how can gufiguer's axioms be correct, and still allow for this type of "metric" to exist?

Well, if you take $d(x,y) = 0$ for all $x,y$, then you indeed get the indiscrete topology. But this is not a metric. So you haven't actually put a metric on the indiscrete topology. If you don't require d(x,y) = 0~\Rightarrow x =y
then you have a thing called a pseudometric. The indiscrete topology is pseudometrizable, but not metrizable. There are many topological spaces however, which are not pseudometrizable.

 

[Jorriss]

When I think about the metric d(x,y) = 0 for all x and y, my only conclusion is that it generates the indiscrete topology. Consider any open ball in this metric space (X, G);

If x and y are distinct points then that is not a metric, it is a pseudometric.

 

[R136a1]

If x and y are distinct points then that is not a metric, it is a pseudometric.

To add something more. A metric is just the Hausdorff version of a pseudometric. So pseudo-metrizable + Hausdorff (or even $T_0$ or $T_1$) implies metrizable. If you don't care about Hausdorff, then pseudometrics are perfectly fine for all purposes.

In general, many concepts in topology have a $T_0$ version and a general version. For example, you have the notion of symmetric spaces, that's the general version. If you add $T_0$ to that, then you get the $T_1$ axiom. Hasudorff is a $T_0$ version, the general version is called preregular, and so on. In the same sense, pseudometrizable is the general version, while the $T_0$ version is metrizable.

 

[1MileCrash]

Ok, so gufiguer's axioms are not for that of a metric, then.

 

[1MileCrash]

To add something more. A metric is just the Hausdorff version of a pseudometric. So pseudo-metrizable + Hausdorff (or even $T_0$ or $T_1$) implies metrizable. If you don't care about Hausdorff, then pseudometrics are perfectly fine for all purposes.

In general, many concepts in topology have a $T_0$ version and a general version. For example, you have the notion of symmetric spaces, that's the general version. If you add $T_0$ to that, then you get the $T_1$ axiom. Hasudorff is a $T_0$ version, the general version is called preregular, and so on. In the same sense, pseudometrizable is the general version, while the $T_0$ version is metrizable.

Wow, that's very cool!

I'm always interested to see what happens when we relax our axioms.

 

[R136a1]

Wow, that's very cool!

I'm always interested to see what happens when we relax our axioms.

You might be interested in this then: http://en.wikipedia.org/wiki/Kolmogorov_quotient#The_Kolmogorov_quotient

 

[Jorriss]

Ok, so gufiguer's axioms are not for that of a metric, then.

No, his axioms are for a metric space but if you remove, say, axiom 3 one arrives at pseudometrics.

 

[1MileCrash]

No, his axioms are for a metric space but if you remove, say, axiom 3 one arrives at pseudometrics.

Right, of course.

Looking at them again, it seems clear to me that his 4 axioms are equivalent to the 3 independent axioms usually given, where his (i) and (iii) together are equivalent to the statement "d(x,y) = 0 iff x = y" when considering the others too.

 

[Jorriss]

Looking at them again, it seems clear to me that his 4 axioms are equivalent to the 3 independent axioms usually given, where his (i) and (iii) together are equivalent to the statement "d(x,y) = 0 iff x = y" when considering the others too.

Yeah, he just opted for more explicit but less compact notation.

 

[gufiguer]

Just continuing. The only metric on $\varnothing$ is $\varnothing$ and the only metric on a singleton X={x} is d={((x,x),0)}, ie, d(x,y)=0, for all x and y in X. If $card(X) \geq 2$, then $\varnothing$ and {((x,x),0)} are not metrics for X as before, but in this case there are (non-countable) infinitely many metrics for X. Prove that as an exercise!