Lipschitz, uniformly, and topologically equivalent metrics
Dave L. Renfro
of problems, on a specific theme, from a topology class
I taught a few years ago.
Corrections and comments are welcome.
TABLE OF CONTENTS
1. LIPSCHITZ EQUIVALENT METRICS
2. UNIFORMLY EQUIVALENT METRICS
3. TOPOLOGICALLY EQUIVALENT METRICS
4. EQUIVALENT METRICS AND SEQUENCES
5. UNIFORMLY EQUIVALENT =/=> LIPSCHITZ EQUIVALENT
6. TOPOLOGICALLY EQUIVALENT =/=> UNIFORMLY EQUIVALENT
7. GENERATING EQUIVALENT METRICS
8. THE BOUNDED METRIC MIN{1, d}
9. THE BOUNDED METRIC d/(1+d)
----------------------
1. LIPSCHITZ EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _Lipschitz_equivalent_ to d' if these conditions
hold.
(L1) The identity function from (X,d) to (X,d') is Lipschitz
continuous and the identity function from (X,d') to
(X,d) is Lipschitz continuous.
(L2) (there exists k > 0)(for all epsilon > 0)
(for all p in X) we have
B(p, k*epsilon) is a subset of B'(p, epsilon)
and
B'(p, k*epsilon) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii epsilon or k*epsilon (as labeled).
(L3) (there exists C > 1)(for all p,q in X) we have
(1/C)*d'(p,q) < d(p,q) < C*d'(p,q).
Remark: If ||*|| and ||*||' are two norms on R^n, then
the metrics they generate are Lipschitz equivalent.
----------------------
2. UNIFORMLY EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _uniformly_equivalent_ to d' if these conditions
hold.
(U1) The identity function from (X,d) to (X,d') is uniformly
continuous and the identity function from (X,d') to
(X,d) is uniformly continuous.
(U2) ( there exists Delta:(0, oo) --> (0, oo) )
(for all epsilon > 0)(for all p in X) we have
B( p, Delta(epsilon) ) is a subset of B'(p, epsilon)
and
B'( p, Delta(epsilon) ) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii epsilon or Delta(epsilon) (as labeled).
(U3) (for all epsilon > 0)(there exists delta > 0)
(for all p in X) we have
B(p, delta) is a subset of B'(p, epsilon)
and
B'(p, delta) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii delta or epsilon (as labeled).
----------------------
3. TOPOLOGICALLY EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _topologically_equivalent_ to d' if these
conditions hold.
(T1) The identity function from (X,d) to (X,d') is
continuous and the identity function from (X,d')
to (X,d) is continuous.
(T2) (for all epsilon > 0)(for all p in X)
(there exists delta > 0) we have
B(p, delta) is a subset of B'(p, epsilon)
and
B'(p, delta) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii delta or epsilon (as labeled).
(T3) (for all subsets U of X) we have
U is open in (X,d) <==> U is open in (X,d').
----------------------
4. EQUIVALENT METRICS AND SEQUENCES
Let d and d' be metrics on X and let {p_n} be a
sequence in X.
(1) Assume d is topologically equivalent to d'.
Prove: p_n --> p in (X,d) if and only if
p_n --> p in (X,d').
(2) Assume d is uniformly equivalent to d'.
Prove: {p_n} is a Cauchy sequence in (X,d)
if and only if {p_n} is a Cauchy
sequence in (X,d').
(3) Assume d is uniformly equivalent to d'.
Prove: (X,d) is complete if and only if
(X,d') is complete.
Remark: (3) shows that completeness is preserved under
uniform equivalence. However, completeness is
also preserved in some cases when d and d'
are not uniformly equivalent. An example is
of this is X = R, d(p,q) = |p - q|, and
d'(p,q) = |p^3 - q^3|. [One of the problems
in the last homework set was to prove d'
is a metric on R. Also, see "7. Generating
Uniformly Equivalent Metrics" below.]
----------------------
5. UNIFORMLY EQUIVALENT =/=> LIPSCHITZ EQUIVALENT
For X = R, define d and d' by d(p,q) = |p - q| and
d'(p,q) = sqrt( |p - q| ).
(1) Prove that d' is a metric on R.
(2) Prove that d is not Lipschitz equivalent to d'.
In fact, prove the following stronger result:
There does not exist C_1 > 0 such that
C_1 * d' < d and there does not exist C_2 > 0
such that d < C_2 * d'.
(3) Prove that d is uniformly equivalent to d'.
----------------------
6. TOPOLOGICALLY EQUIVALENT =/=> UNIFORMLY EQUIVALENT
For X = R, define d and d' by d(p,q) = |p - q| and
d'(p,q) = | g(p) - g(q) |, where g:R --> (-1,1) is
given by g(t) = t / (1 + |t|).
(1) Prove that d' is a metric on R.
(2) Prove that g : (R,d') --> ( (-1,1), d ) is an
isometry, and hence (R,d') is not complete.
Using an earlier problem, explain why this
shows d is not uniformly equivalent to d'.
(3) Prove that d is topologically equivalent to d'.
Hint: Use the continuity of g to show that
id: (R,d) --> (R,d') is continuous, and
use the continuity of g^(-1) to show that
id:(R,d') --> (R,d) is continuous.
Note: For notational convenience, I am using g,
which was originally defined on the *set* R of
real numbers, to denote the function defined on
the *metric space* (R,d').
----------------------
7. GENERATING EQUIVALENT METRICS
Let (X, d1) and (Y, d2) be metric spaces, f:X --> Y
be a bijection, and define d: X x X --> [0, oo) by
d(p,q) = d2( f(p), f(q) ).
(1) Prove: d is a metric on X.
(2) Prove: d is topologically equivalent to d1
<==> f is a homeomorphism.
Remark: See Willard's "General Topology" [2C(4,5), p. 20]
for pseudo-metric space versions of (1) & (2).
(3) Prove: d is uniformly equivalent to d1
<==> f is a uniform homeomorphism.
(4) Prove: d is Lipschitz equivalent to d1
<==> f is bi-Lipschitz.
(5) Prove: d = d1 <==> f is an isometry.
The results above involve defining a distance function
by applying a certain function to two points and then
applying a certain metric to the resulting two values.
We can also define a distance function by applying a
certain metric to two points and then applying a certain
function to the resulting value.
(6) Let h:[0, oo) --> [0, oo) be continuous, non-increasing,
subadditive, and such that h(p) = 0 <==> p = 0. If
(X,d) is a metric space, define d':X x X --> [0, oo)
by d'(p,q) = h( d(p,q) ). Then d' is a metric on X
and d' is uniformly equivalent to d. [For a proof,
see Theorem 16.12 (p. 176) in Section 16.9 of William
G. Faris's book "Real Analysis Structures"
<http://math.arizona.edu/~faris/realweb/real.pdf>.]
Remark: Several papers have been written that study,
from a real analysis viewpoint, the properties
of functions h:[0, oo) --> [0, oo) that can be
used to define a metric d' from a given metric d
via d'(p,q) = h( d(p,q) ).
http://www.google.com/search?q=metric-preserving-functions
----------------------
8. THE BOUNDED METRIC MIN{1, d}
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = min{1, d}.
(1) Prove: d' is a metric on X.
(2) Prove: d' is uniformly equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not Lipschitz equivalent to d.
Challenge: Can you find an easy to verify condition on
(X,d) that is both necessary and sufficient
for d' to be Lipschitz equivalent to d?
----------------------
9. THE BOUNDED METRIC d/(1+d)
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = d/(1+d).
(1) Prove: d' is a metric on X.
(2) Prove: d' is topologically equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not uniformly equivalent to d.
Hint: Use 7(3) above with X = [0, oo), d1(p,q) = |p-q|,
Y = [0,1), d2 is the restriction of d1 to Y x Y,
and f(x) = x/(1+x).
----------------------
Dave L. Renfro
David C. Ullrich
wrote:
>What follows is another ASCII version of a collection
>of problems, on a specific theme, from a topology class
>I taught a few years ago.
>
>Corrections and comments are welcome.
>
>[...]
>8. THE BOUNDED METRIC MIN{1, d}
>9. THE BOUNDED METRIC d/(1+d)
You might note that 8 and 9 are special cases of
one construction: If d is a metric and phi satisfies
a certain list of natural conditions then phi \circ d
is a metric.
I think it's phi(0) = 0, phi(x) > 0 for all x > 0,
phi non-decreasing and concave... right, that makes
phi\circ d a metric, and if phi is continuous at 0
then the new metric is uniformly equivalent to the
original.
************************
David C. Ullrich
Dave L. Renfro
>> [...]
>> 8. THE BOUNDED METRIC MIN{1, d}
>> 9. THE BOUNDED METRIC d/(1+d)
David C. Ullrich wrote:
> You might note that 8 and 9 are special cases of
> one construction: If d is a metric and phi satisfies
> a certain list of natural conditions then phi \circ d
> is a metric.
>
> I think it's phi(0) = 0, phi(x) > 0 for all x > 0,
> phi non-decreasing and concave... right, that makes
> phi\circ d a metric, and if phi is continuous at 0
> then the new metric is uniformly equivalent to the
> original.
That was what "7. GENERATING EQUIVALENT METRICS" dealt
with, specifically 7(6). The google search I gave
(phrase = "metric preserving functions") takes you
to several papers dealing with this (one of which I
was a referee for). I'm thinking now that it would
be better to put these two specific examples before
this topic. I need to correct an error I noticed after
I made my post yesterday, anyway, as well as make some
tweaking of the wording in a few places. I noticed the
error while I was incorporating some minor changes in
wording on my original LaTex document, improvements that
I thought of while I was writing the post (and naturally,
while I was putting these changes into my original LaTex
document, I came up with some additional changes in wording
that I want to make in the post), I noticed I said the phi
function was non-increasing in the post, not non-decreasing.
By the way, I wanted to reply in the L^p for 0 < p < 1
thread and in the Holder continuity thread, but I was
too tired of doing this stuff by the time I finished
with all that metric equivalence stuff yesterday.
I'll get to those shortly.
Dave L. Renfro
Dave L. Renfro
post ["non-increasing" in 7(6) should be
"non-decreasing], along with a few other changes.
What follows is another ASCII version of a collection
of problems, on a specific theme, from a topology class
I taught a few years ago.
Corrections and comments are welcome.
TABLE OF CONTENTS
1. LIPSCHITZ EQUIVALENT METRICS
2. UNIFORMLY EQUIVALENT METRICS
3. TOPOLOGICALLY EQUIVALENT METRICS
4. EQUIVALENT METRICS AND SEQUENCES
5. UNIFORMLY EQUIVALENT =/=> LIPSCHITZ EQUIVALENT
6. TOPOLOGICALLY EQUIVALENT =/=> UNIFORMLY EQUIVALENT
7. THE BOUNDED METRIC MIN{1, d}
8. THE BOUNDED METRIC d/(1+d)
9. GENERATING EQUIVALENT METRICS
----------------------
1. LIPSCHITZ EQUIVALENT METRICS
----------------------
2. UNIFORMLY EQUIVALENT METRICS
hold. (Warning: Some authors use the term "uniformly
equivalent metrics" to mean what I'm calling "Lipschitz
equivalent metrics.)
(U1) The identity function from (X,d) to (X,d') is uniformly
continuous and the identity function from (X,d') to
(X,d) is uniformly continuous.
(U2) ( there exists Delta:(0, oo) --> (0, oo) )
(for all epsilon > 0)(for all p in X) we have
B( p, Delta(epsilon) ) is a subset of B'(p, epsilon)
and
B'( p, Delta(epsilon) ) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii epsilon or Delta(epsilon) (as labeled).
(U3) (for all epsilon > 0)(there exists delta > 0)
(for all p in X) we have
B(p, delta) is a subset of B'(p, epsilon)
and
B'(p, delta) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii delta or epsilon (as labeled).
Remark: Problem 35G (p. 244) in Willard's "General
Topology" says that Lipschitz equivalent
metrics on a set X generate the same uniformity
on X. Exercise 8.1.A(a) (p. 535) in Engelking's
"General Topology" (1977 edition) says that
two metrics on a set X generate the same
uniformity on X <==> the metrics are
uniformly equivalent.
----------------------
3. TOPOLOGICALLY EQUIVALENT METRICS
----------------------
of this is X = R, d(p,q) = |p - q|, and
d'(p,q) = |p^3 - q^3|. [One of the problems
in the last homework set was to prove d'
is a metric on R. Also, see "9. Generating
Equivalent Metrics" below.]
----------------------
----------------------
real numbers, to also denote the function defined
on the *metric space* (R,d').
----------------------
7. THE BOUNDED METRIC MIN{1, d}
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = min{1, d}.
(1) Prove: d' is a metric on X.
(2) Prove: d' is uniformly equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not Lipschitz equivalent to d.
Challenge: Can you find an easy to verify condition on
(X,d) that is both necessary and sufficient
for d' to be Lipschitz equivalent to d?
----------------------
8. THE BOUNDED METRIC d/(1+d)
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = d/(1+d).
(1) Prove: d' is a metric on X.
(2) Prove: d' is topologically equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not uniformly equivalent to d.
Hint: Use 9(3) below with X = [0, oo), d1(p,q) = |p-q|,
Y = [0,1), d2 is the restriction of d1 to Y x Y,
and f(x) = x/(1+x).
----------------------
9. GENERATING EQUIVALENT METRICS
Let (X, d1) and (Y, d2) be metric spaces, f:X --> Y
be a bijection, and define d: X x X --> [0, oo) by
d(p,q) = d2( f(p), f(q) ).
(1) Prove: d is a metric on X.
(2) Prove: d is topologically equivalent to d1
<==> f is a homeomorphism.
Remark: See Willard's "General Topology" [2C(4,5), p. 20]
for pseudo-metric space versions of (1) & (2).
(3) Prove: d is uniformly equivalent to d1
<==> f is a uniform homeomorphism.
(4) Prove: d is Lipschitz equivalent to d1
<==> f is bi-Lipschitz.
(5) Prove: d = d1 <==> f is an isometry.
The results above involve defining a distance function
by applying a certain function to two points and then
applying a certain metric to the resulting two values.
We can also define a distance function by applying a
certain metric to two points and then applying a certain
function to the resulting value. Note that #7 and #8
above are special cases of this second construction.
(6) Let h:[0, oo) --> [0, oo) be continuous, non-decreasing,
subadditive, and such that h(t) = 0 <==> t = 0. Let
(X,d) be a metric space and define d':X x X --> [0, oo)
by d'(p,q) = h( d(p,q) ). Then d' is a metric on X
and d' is uniformly equivalent to d. [This particular
version can be found in William G. Faris's book "Real
Analysis Structures" (Section 16.9, Theorem 16.12, p. 176)
<http://math.arizona.edu/~faris/realweb/real.pdf>.]
Remark: Several papers have been written that study,
from a real analysis viewpoint, the properties
of functions h:[0, oo) --> [0, oo) that can be
used to define a metric d' from a given metric d
via d'(p,q) = h( d(p,q) ).
http://www.google.com/search?q=metric-preserving-functions
----------------------
Dave L. Renfro
William Elliot
> What follows is another ASCII version of a collection
> of problems, on a specific theme, from a topology class
> I taught a few years ago.
>
> Corrections and comments are welcome.
Didn't notice any of these constructions of metrics
from metrics in your well written synopsis.
The sup of a collection of metrics is a metric.
the inf of a collection of metrics is a pseudo-metric.
and the min a collection of metrics is a metric.
Is there any preservation of topological equivalence?
Let S = { x,y } and dj(x,y) = 1/j indicate a metric.
Then inf{ dj | j in N } is a genuine pseudo-metric.
----
David C. Ullrich
wrote:
>Dave L. Renfro wrote (in part):
>
>>> [...]
>>> 8. THE BOUNDED METRIC MIN{1, d}
>>> 9. THE BOUNDED METRIC d/(1+d)
>
>David C. Ullrich wrote:
>
>> You might note that 8 and 9 are special cases of
>> one construction: If d is a metric and phi satisfies
>> a certain list of natural conditions then phi \circ d
>> is a metric.
>>
>> I think it's phi(0) = 0, phi(x) > 0 for all x > 0,
>> phi non-decreasing and concave... right, that makes
>> phi\circ d a metric, and if phi is continuous at 0
>> then the new metric is uniformly equivalent to the
>> original.
>
>That was what "7. GENERATING EQUIVALENT METRICS" dealt
>with, specifically 7(6).
Oh. Yes it is - never mind...
>The google search I gave
>(phrase = "metric preserving functions") takes you
>to several papers dealing with this (one of which I
>was a referee for). I'm thinking now that it would
>be better to put these two specific examples before
>this topic. I need to correct an error I noticed after
>I made my post yesterday, anyway, as well as make some
>tweaking of the wording in a few places. I noticed the
>error while I was incorporating some minor changes in
>wording on my original LaTex document, improvements that
>I thought of while I was writing the post (and naturally,
>while I was putting these changes into my original LaTex
>document, I came up with some additional changes in wording
>that I want to make in the post), I noticed I said the phi
>function was non-increasing in the post, not non-decreasing.
>
>By the way, I wanted to reply in the L^p for 0 < p < 1
>thread and in the Holder continuity thread, but I was
>too tired of doing this stuff by the time I finished
>with all that metric equivalence stuff yesterday.
>I'll get to those shortly.
>
>Dave L. Renfro
************************
David C. Ullrich
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Sun, 17 Sep 2006, Dave L. Renfro wrote:
>
>> What follows is another ASCII version of a collection
>> of problems, on a specific theme, from a topology class
>> I taught a few years ago.
>>
>> Corrections and comments are welcome.
>
>Didn't notice any of these constructions of metrics
>from metrics in your well written synopsis.
>
>The sup of a collection of metrics is a metric.
>the inf of a collection of metrics is a pseudo-metric.
How do you prove that?
>and the min a collection of metrics is a metric.
Presumably you mean the min of a finite collection
of metrics, since an infinite collection need not
have a min.
Anyway, say X = {a,b,c}, define
d1(a,b) = 1, d1(b,c) = 3, d1(a,c) = 4
d2(a,b) = 3, d2(b,c) = 1, d2(a,c) = 4.
Then d = min(d1, d2) is not a metric.
You just make this stuff up, right?
>Is there any preservation of topological equivalence?
>
>Let S = { x,y } and dj(x,y) = 1/j indicate a metric.
>
>Then inf{ dj | j in N } is a genuine pseudo-metric.
>
>----
************************
David C. Ullrich
William Elliot
> <ma...@hevanet.remove.com> wrote:
> >On Sun, 17 Sep 2006, Dave L. Renfro wrote:
> >
> >Didn't notice any of these constructions of metrics
> >from metrics in your well written synopsis.
> >
> >The sup of a collection of metrics is a metric.
The sup of a bounded above collection of metrics is a metric.
> >the inf of a collection of metrics is a pseudo-metric.
> How do you prove that?
>
> >and the min a collection of metrics is a metric.
>
The inf of a bounded below collections of metrics is a metric.
> Presumably you mean the min of a finite collection
> of metrics, since an infinite collection need not
> have a min.
>
> Anyway, say X = {a,b,c}, define
>
> d1(a,b) = 1, d1(b,c) = 3, d1(a,c) = 4
> d2(a,b) = 3, d2(b,c) = 1, d2(a,c) = 4.
>
> Then d = min(d1, d2) is not a metric.
>
Ok, I found where my reasoning when wrong.
I wonder if there's a work around for that
with the sup proposition (as amended).
> You just make this stuff up, right?
>
So did Cantor. ;-)
No, I got the idea from the discussion with rusty.
Have you counter example for the amended sup proposition?
Anyway, before leaping to generalizations, I should have
detailed a proof that the max of two metrics is a metric
as rusty asserted. Later.
Dave L. Renfro
> Ok, I found where my reasoning when wrong.
> I wonder if there's a work around for that
> with the sup proposition (as amended).
Below are a couple of papers that relate to the
kinds of issues you're bringing up. Also, if you
ever make it to a university library, Eduard Cech's
1966 book "Topological Spaces" (revised and translated
by Zdenek Frolek and Miroslav Katetov) is a gold mine
of lattice-theoretic results for various collections
of generalized metrics and generalized topologies on
a given set.
Merrill E. Shanks, "The space of metrics on a compact
metrizable space", American Journal of Mathematics
66 (1944), 461-469. [Warning: Shanks uses the term
"quasi-metric" for what everyone else (now, at least)
calls "pseudo-metric".] [MR 6,95c; Zbl 63.06939]
http://www.emis.de/cgi-bin/MATH-item?0063.06939
M. Katetov, "Metrics on an arc", Studia Mathematica
31 (1968), 547-554. [MR 41 #7624; Zbl 181.26004]
http://www.emis.de/cgi-bin/MATH-item?0181.26004
Dave L. Renfro
William Elliot
Newsgroups: sci.math
Subject: Re: Lipschitz, uniformly, and topologically equivalent metrics
William Elliot wrote (in part):
> Ok, I found where my reasoning when wrong.
> I wonder if there's a work around for that
> with the sup proposition (as amended).
> Below are a couple of papers that relate to the
> kinds of issues you're bringing up. Also, if you
> ever make it to a university library, Eduard Cech's
> 1966 book "Topological Spaces" (revised and translated
> by Zdenek Frolek and Miroslav Katetov) is a gold mine
> of lattice-theoretic results for various collections of
> generalized metrics and generalized topologies on a given set.
Let M = M(S) be the set of all metrics for S, ordered pointwise.
M(S) is a bounded complete lattice.
Let C be a bounded above subset of M. Then D = sup C in M
sup C exists since for all x,y, { d(x,y) | d in C } is bounded above.
D(x,y) = 0 iff x = y
D(x,y) = D(y,x)
D(x,z) = sup{ d(x,z) | d in C }
<= sup{ d(x,y) + d(y,z) | d in C }
<= sup{ d(x,y) | d in C } + sup{ d(y,z) | d in C }
= D(x,y) + D(y,z)
Lemma. inf_j aj + inf_j bj <= inf_j (aj + bj)
sup_j (aj + bj) <= sup_j aj + sup_j bj
One notes that an upper bound d for C
isn't necessarly a bounded metric
That the metrics d_n(x,y) = 1/n when x /= y,
aren't bounded below (when S is multi-point).
> Merrill E. Shanks, "The space of metrics on a compact
> metrizable space", American Journal of Mathematics
> 66 (1944), 461-469. [Warning: Shanks uses the term
> "quasi-metric" for what everyone else (now, at least)
> calls "pseudo-metric".] [MR 6,95c; Zbl 63.06939]
> http://www.emis.de/cgi-bin/MATH-item?0063.06939
> M. Katetov, "Metrics on an arc", Studia Mathematica
> 31 (1968), 547-554. [MR 41 #7624; Zbl 181.26004]
> http://www.emis.de/cgi-bin/MATH-item?0181.26004
I've already more papers upon the topic of the topology
of (partially) ordered sets, than I have time to read.
----
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Mon, 18 Sep 2006, David C. Ullrich wrote:
>> <ma...@hevanet.remove.com> wrote:
>> >On Sun, 17 Sep 2006, Dave L. Renfro wrote:
>> >
>> >Didn't notice any of these constructions of metrics
>> >from metrics in your well written synopsis.
>> >
>> >The sup of a collection of metrics is a metric.
>
>The sup of a bounded above collection of metrics is a metric.
>
>> >the inf of a collection of metrics is a pseudo-metric.
>
>> How do you prove that?
>>
>> >and the min a collection of metrics is a metric.
>>
>The inf of a bounded below collections of metrics is a metric.
Nope.
>> Presumably you mean the min of a finite collection
>> of metrics, since an infinite collection need not
>> have a min.
>>
>> Anyway, say X = {a,b,c}, define
>>
>> d1(a,b) = 1, d1(b,c) = 3, d1(a,c) = 4
>> d2(a,b) = 3, d2(b,c) = 1, d2(a,c) = 4.
>>
>> Then d = min(d1, d2) is not a metric.
>>
>Ok, I found where my reasoning when wrong.
>I wonder if there's a work around for that
>with the sup proposition (as amended).
>
>> You just make this stuff up, right?
>>
>So did Cantor. ;-)
Uh, that's not true either.
>No, I got the idea from the discussion with rusty.
>Have you counter example for the amended sup proposition?
Of course not - the proof of that is trivial.
>Anyway, before leaping to generalizations,
Before stating things as facts you should make certain
you know how to prove them.
>I should have
>detailed a proof that the max of two metrics is a metric
>as rusty asserted. Later.
>
>> >Is there any preservation of topological equivalence?
>> >
>> >Let S = { x,y } and dj(x,y) = 1/j indicate a metric.
>> >
>> >Then inf{ dj | j in N } is a genuine pseudo-metric.
************************
David C. Ullrich
William Elliot
> >Ok, I found where my reasoning when wrong.
> >I wonder if there's a work around for that
> >with the sup proposition (as amended).
See Dave's reply to this and my reply to him.
> Of course not - the proof of that is trivial.
>
Then I intuited a correct conjecture.
> >Anyway, before leaping to generalizations,
>
> Before stating things as facts you should make certain
> you know how to prove them.
>
I just did, that the pointwise ordered set of metrics for a space is a
bounded complete lattice which thus establishes every bounded below set of
metrics has an inf metric (which of course isn't the pointwise inf).
Care to go the next step and relate some one of topologies for this
lattice to the topology given by the metric for the space of metrics as
given by Dave in another thread? Anyway, have you any comments about
M(S)/~ as I constructed there?
David C. Ullrich
<ma...@hevanet.remove.com> wrote:
>On Tue, 19 Sep 2006, David C. Ullrich wrote:
>
>> >Ok, I found where my reasoning when wrong.
>> >I wonder if there's a work around for that
>> >with the sup proposition (as amended).
>
>See Dave's reply to this and my reply to him.
>
>> Of course not - the proof of that is trivial.
Huh? I don't know why you're lying about this.
That was my response to something else you said:
"
>Have you counter example for the amended sup proposition?
Of course not - the proof of that is trivial.
"
>Then I intuited a correct conjecture.
>
>> >Anyway, before leaping to generalizations,
>>
>> Before stating things as facts you should make certain
>> you know how to prove them.
>>
>I just did, that the pointwise ordered set of metrics for a space is a
>bounded complete lattice which thus establishes every bounded below set of
>metrics has an inf metric (which of course isn't the pointwise inf).
>
>Care to go the next step and relate some one of topologies for this
>lattice to the topology given by the metric for the space of metrics as
>given by Dave in another thread? Anyway, have you any comments about
>M(S)/~ as I constructed there?
************************
David C. Ullrich
Dave L. Renfro
did not appear in the Sept. 16 & 17 versions.
What follows is another ASCII version of a collection
of problems, on a specific theme, from a topology class
I taught a few years ago.
Corrections and comments are welcome.
TABLE OF CONTENTS
1. LIPSCHITZ EQUIVALENT METRICS
2. UNIFORMLY EQUIVALENT METRICS
3. TOPOLOGICALLY EQUIVALENT METRICS
4. EQUIVALENT METRICS AND SEQUENCES
5. UNIFORMLY EQUIVALENT =/=> LIPSCHITZ EQUIVALENT
6. TOPOLOGICALLY EQUIVALENT =/=> UNIFORMLY EQUIVALENT
7. THE BOUNDED METRIC MIN{1, d}
8. THE BOUNDED METRIC d/(1+d)
9. GENERATING EQUIVALENT METRICS
10. REFERENCES
----------------------
1. LIPSCHITZ EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _Lipschitz_equivalent_ to d' if these conditions
hold.
(L1) The identity function from (X,d) to (X,d') is Lipschitz
continuous and the identity function from (X,d') to
(X,d) is Lipschitz continuous.
(L2) (there exists k > 0)(for all epsilon > 0)
(for all p in X) we have
B(p, k*epsilon) is a subset of B'(p, epsilon)
and
B'(p, k*epsilon) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii epsilon or k*epsilon (as labeled).
(L3) (there exists C > 1)(for all p,q in X) we have
(1/C)*d'(p,q) < d(p,q) < C*d'(p,q).
Remark: If ||*|| and ||*||' are two norms on R^n,
then the metrics they generate are Lipschitz
equivalent. [This is a well known result in
functional analysis, although it's often
stated in a weaker form, with "uniformly
or "topological" in place of "Lipschitz".]
----------------------
2. UNIFORMLY EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _uniformly_equivalent_ to d' if these conditions
hold. (Warning: Some authors use the term "uniformly
equivalent metrics" to mean what I'm calling "Lipschitz
equivalent metrics.)
(U1) The identity function from (X,d) to (X,d') is uniformly
continuous and the identity function from (X,d') to
(X,d) is uniformly continuous.
(U2) ( there exists Delta:(0, oo) --> (0, oo) )
(for all epsilon > 0)(for all p in X) we have
B( p, Delta(epsilon) ) is a subset of B'(p, epsilon)
and
B'( p, Delta(epsilon) ) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii epsilon or Delta(epsilon) (as labeled).
(U3) (for all epsilon > 0)(there exists delta > 0)
(for all p in X) we have
B(p, delta) is a subset of B'(p, epsilon)
and
B'(p, delta) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii delta or epsilon (as labeled).
Remark: Problem 35G (p. 244) in Willard [6] says that
Lipschitz equivalent metrics on a set X generate
the same uniformity on X. Exercise 8.1.A(a) (p. 535)
in Engelking [1] says that two metrics on a set X
generate the same uniformity on X <==> the metrics
are uniformly equivalent.
----------------------
3. TOPOLOGICALLY EQUIVALENT METRICS
Let d and d' be two metrics on a set X. Prove that the
following conditions are logically equivalent. We say
that d is _topologically_equivalent_ to d' if these
conditions hold.
(T1) The identity function from (X,d) to (X,d') is
continuous and the identity function from (X,d')
to (X,d) is continuous.
(T2) (for all epsilon > 0)(for all p in X)
(there exists delta > 0) we have
B(p, delta) is a subset of B'(p, epsilon)
and
B'(p, delta) is a subset of B(p, epsilon),
where B, B' are open balls in (X,d), (X,d') with
center p and radii delta or epsilon (as labeled).
(T3) (for all subsets U of X) we have
U is open in (X,d) <==> U is open in (X,d').
Remark 1: If d and d' are topologically equivalent
metrics on X and X is compact under the
topology that these metrics induce, then d
is uniformly equivalent to d'. [This follows
immediately from (U1) and (T1) above.]
Remark 2: Levine [3] proved that if f : (X1,d1) --> (X2,d2)
is continuous, then there exists a metric d1*
on X1 that is topologically equivalent to d1
such that f: (X1,d1*) --> (X2,d2) is uniformly
continuous. In fact, such a metric is given by
d1*(p,q) = d1(p,q) + d2( f(p), f(q) ). Levine [3]
goes on to show that such a metric d1* can be
found that simultaneously "converts" any specified
finite collection of continuous functions into
uniformly continuous functions.
Levine [3] then gives an example to show that
there exists a continuous function from (X1,d1)
to (X2,d2) that cannot be made uniformly
continuous by any replacement of d2 by a
metric that is topologically equivalent to d2.
Finally, Levine [3] proves that given any
countable collection of continuous functions
from (X1,d1) to (X2,d2), there exists metrics
d1* and d2*, topologically equivalent to
d1 and d2, such that each of these functions
are made uniformly continuous when viewed
as functions from (X1,d1*) to (X2,d2*).
For related results, see Gray/Hejcman [2],
Levine [4], and Snipes [5].
----------------------
4. EQUIVALENT METRICS AND SEQUENCES
Let d and d' be metrics on X and let {p_n} be a
sequence in X.
(1) Assume d is topologically equivalent to d'.
Prove: p_n --> p in (X,d) if and only if
p_n --> p in (X,d').
(2) Assume d is uniformly equivalent to d'.
Prove: {p_n} is a Cauchy sequence in (X,d)
if and only if {p_n} is a Cauchy
sequence in (X,d').
(3) Assume d is uniformly equivalent to d'.
Prove: (X,d) is complete if and only if
(X,d') is complete.
Remark: (3) shows that completeness is preserved under
uniform equivalence. However, completeness is
also preserved in some cases when d and d'
are not uniformly equivalent. An example
of this is X = R, d(p,q) = |p - q|, and
d'(p,q) = |p^3 - q^3|. [One of the problems
in the last homework set was to prove d'
is a metric on R. Also, see "9. Generating
Equivalent Metrics" below.]
----------------------
----------------------
real numbers, to also denote the function defined
on the *metric space* (R,d').
----------------------
7. THE BOUNDED METRIC MIN{1, d}
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = min{1, d}.
(1) Prove: d' is a metric on X.
(2) Prove: d' is uniformly equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not Lipschitz equivalent to d.
Challenge: Can you find an easy to verify condition on
(X,d) that is both necessary and sufficient
for d' to be Lipschitz equivalent to d?
----------------------
8. THE BOUNDED METRIC d/(1+d)
Let (X,d) be a metric space and define d':X x X --> [0, oo)
by d' = d/(1+d).
(1) Prove: d' is a metric on X.
(2) Prove: d' is topologically equivalent to d.
(3) Prove: Give a specific _example_ in which d'
is not uniformly equivalent to d.
Hint: Use 9(3) below with X = [0, oo), d1(p,q) = |p-q|,
Y = [0,1), d2 is the restriction of d1 to Y x Y,
and f(x) = x/(1+x).
----------------------
9. GENERATING EQUIVALENT METRICS
Let (X, d1) and (Y, d2) be metric spaces, f:X --> Y
be a bijection, and define d: X x X --> [0, oo) by
d(p,q) = d2( f(p), f(q) ).
(1) Prove: d is a metric on X.
(2) Prove: d is topologically equivalent to d1
<==> f is a homeomorphism.
Remark: See Willard [6] [2C(4,5), p. 20] for
pseudo-metric space versions of (1) & (2).
(3) Prove: d is uniformly equivalent to d1
<==> f is a uniform homeomorphism.
(4) Prove: d is Lipschitz equivalent to d1
<==> f is bi-Lipschitz.
(5) Prove: d = d1 <==> f is an isometry.
The results above involve defining a distance function
by applying a certain function to two points and then
applying a certain metric to the resulting two values.
We can also define a distance function by applying a
certain metric to two points and then applying a certain
function to the resulting value. Note that #7 and #8
above are special cases of this second construction.
(6) Let h:[0, oo) --> [0, oo) be continuous, non-decreasing,
subadditive, and such that h(t) = 0 <==> t = 0. Let
(X,d) be a metric space and define d':X x X --> [0, oo)
by d'(p,q) = h( d(p,q) ). Then d' is a metric on X
and d' is uniformly equivalent to d. [This particular
version can be found in William G. Faris's book "Real
Analysis Structures" (Section 16.9, Theorem 16.12, p. 176)
<http://math.arizona.edu/~faris/realweb/real.pdf>.]
Remark: Several papers have been written that study,
from a real analysis viewpoint, the properties
of functions h:[0, oo) --> [0, oo) that can be
used to define a metric d' from a given metric d
via d'(p,q) = h( d(p,q) ).
http://www.google.com/search?q=metric-preserving-functions
----------------------
10. REFERENCES
[1] Ryszard Engelking, GENERAL TOPOLOGY, Monografie
Matematyczne #60, PWN-Polish Scientific Publishers,
1977, 626 pages. [MR 58 #18316b; Zbl 373.54002]
http://www.emis.de/cgi-bin/MATH-item?0373.54002
[2] Peter W. Gray and Jan Hejcman, "Solution to Monthly
Problem #5515", American Mathematical Monthly 75 #7
(August/September 1968), 797-798. [Proposed by John
W. Wyman, AMM 74 (1967), p. 872.]
[3] Norman Levine, "Remarks on uniform continuity in
metric spaces", American Mathematical Monthly 67 #6
(June/July 1960), 562-563. [MR 22 #7105; Zbl 100.18602]
http://www.emis.de/cgi-bin/MATH-item?0100.18602
[4] Norman Levine, "A note on the Lipschitz condition
in metric spaces", Proceedings of the American
Mathematical Society 13 #2 (April 1962), 314-315.
[MR 24 #A2935; Zbl 108.35603]
http://www.emis.de/cgi-bin/MATH-item?0108.35603
[5] Ray F. Snipes, "Is every continuous function uniformly
continuous?", Mathematics Magazine 57 #3 (May 1984),
169-173. [MR 85d:26002; Zbl 549.26002]
http://www.emis.de/cgi-bin/MATH-item?0549.26002
[6] Stephen Willard, GENERAL TOPOLOGY, Addison-Wesley, 1970,
xii + 369 pages. [Reprinted by Dover Publications in 2004.]
[MR 41 #9173; Zbl 205.26601]
http://www.emis.de/cgi-bin/MATH-item?0205.26601
----------------------
Dave L. Renfro