Edwin Koh

Posted on Friday, 05 March, 2004 - 06:34 pm:

Suppose

d

and

D

are two metrics defined on a set

S

which is compact with respect to

D

. Suppose

\forall (s, t) \in S \times S

, we have

d (s, t) \leq D (s, t)

. Are

d

and

D

necessarily uniformly equivalent? i.e. is there a constant

C

such that

D (s, t) \leq C d (s, t)

for all

(s, t) \in S \times S

Michael Doré

Posted on Friday, 05 March, 2004 - 08:34 pm:

Try letting S = {0} u {1/n | n > 0 is an integer}. Take d and D to be British rail metrics with London at 0.

[A British rail metric on a set S is a metric with a point L (London) such that:

d(x,y) = d(x,L) + d(L,y)

for all distinct x,y.

It is called a British rail metric because the railway distance between any two cities in England is the distance from the first city to London and then to the second city. In our example L = 0. Note that to specify a British rail metric you only need to specify d(x,L) for each x =/= L, and these can be chosen to be any positive numbers.]

It is straightfoward to see that (S,D) is compact, as long as D(0,1/n) -> 0 as n-> infinity. But it is easy to arrange d < = D and d(0,1/n)/D(0,1/n) -> 0 as n-> infinity. For example pick d(0,1/n) = 1/n² and D(0,1/n) = 1/n. This is thus a counterexample.

Edwin Koh

Posted on Sunday, 21 March, 2004 - 04:16 pm:

If S is the Riemann sphere and D is the spherical metric, is it still possible to find a counterexample?

Michael Doré

Posted on Sunday, 21 March, 2004 - 07:10 pm:

Yes. In fact I think I can show that given any metric space

(X, D)

with a non-isolated point 0 it is possible to find another metric

d

X

with

d \leq D

such that

d (0, x) / D (0, x)

can be made arbitrarily small.

Proof: Define $e (x, y) = min (D (x, 0) + D (y, 0), 1) \times D (x, y)$ . Now $e (x, y) = 0 \Leftrightarrow x = y$ and $e (x, y) = e (y, x)$ . The only problem is that $e$ needn't satisfy the triangle inequality. However that is easy to amend: define $d (x, y) = inf_{x_{0}, \dots, x_{n}} \sum_{r = 0}^{n - 1} e (x_{r}, x_{r + 1})$ where the inf is over all finite sequences $x_{0}$ , ..., $x_{n}$ with $x_{0} = x$ , $x_{n} = y$ . Clearly $d$ is symmetric, satisfies the triangle inequality and $d (x, x) = 0$ . We need to check that $d (x, y) = 0 \Rightarrow x = y$ .

Suppose WLOG that $x \neq y$ and $y \neq 0$ - we want $d (x, y) > 0$ . Given a sequence $x = x_{0}$ , ..., $x_{n} = y$ then if there is no $r$ with $D (x_{r}, 0) < D (y, 0) / 2$ then $e (x_{r}, x_{r + 1}) \geq min (D (y, 0), 1) D (x_{r}, x_{r + 1})$ so $\sum_{r = 0}^{n - 1} e (x_{r}, x_{r + 1}) \geq min (D (y, 0), 1) \sum_{r = 0}^{n - 1} D (x_{r}, x_{r + 1}) \geq min (D (y, 0), 1) D (x, y)$ .

Otherwise let $s$ be maximal such that $D (x_{s}, 0) < D (y, 0) / 2$ . Then $\sum_{r = s}^{n - 1} e (x_{r}, x_{r + 1}) \geq \sum_{r = s}^{n - 1} min (d (0, x_{r + 1}), 1) D (x_{r}, x_{r + 1}) \geq \sum_{r = s}^{n - 1} D (y, 0) / 2 D (x_{r}, x_{r + 1}) \geq D (y, 0) / 2 D (x_{s}, y) \geq D (y, 0) / 2 (D (y, 0) - D (x_{s}, 0)) > D (y, 0) / 2 \times D (y, 0) / 2$ .

Taking the inf, we see $d (x, y) \geq min (min (D (y, 0), 1) D (x, y), (D (y, 0) / 2)^{2}) > 0$ as required.

So $d$ is a metric and $d \leq e \leq D$ . Let $x_{n}$ be a sequence with $D (x_{n}, 0) \to 0$ . Then if $D (x_{n}, 0) \leq 1$ we have $e (x_{n}, 0) = D (x_{n}, 0)^{2}$ so $d (x_{n}, 0) / D (x_{n}, 0) \leq D (x_{n}, 0) \to 0$ so we're done.