How can one prove that even inequivalent metrics can
result in the same class of borel sets?

Would it be sufficient to take two different metrics
both spanning over [0;1]?

On Wed, 24 Jan 2007, The Cool Giraffe wrote:



How can one prove that even inequivalent metrics can
result in the same class of borel sets?



You can't. Consider the discrete metric and the usual metric for R.

Would it be sufficient to take two different metrics
both spanning over [0;1]?

William Elliot wrote:
On Wed, 24 Jan 2007, The Cool Giraffe wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?

You can't. Consider the discrete metric and the usual metric for R.

Ignore WE's faulty logic here.

Would it be sufficient to take two different metrics
both spanning over [0;1]?

You need to find an example of two inequivalent metrics on some set.
Show that the open sets in each topology generate the same
sigma-field. For example, build the sigma-fields hierarchically. If
at some point you have the same collection of sets in each case, you
are done.

Stephen J. Herschkorn wrote/skrev/kaita/popisal/schreibt :




On Wed, 24 Jan 2007, The Cool Giraffe wrote:



How can one prove that even inequivalent metrics can
result in the same class of borel sets?



Would it be sufficient to take two different metrics
both spanning over [0;1]?


You need to find an example of two inequivalent metrics on some set.
Show that the open sets in each topology generate the same
sigma-field. For example, build the sigma-fields hierarchically. If
at some point you have the same collection of sets in each case, you
are done.




So, how about the following set up. I understand it's far
from the best approach but i'm trying my best.

first metric
min( sup( |x(t)-y(t)| ) ; 6) , both sup and min for t in [0;1]

second metric
min( inf( |x(t)-y(t)| ) ; 6) , both inf and min for t in [0;1]

Both x(t) and y(t) are continuous functions on [0;1].

The way i see it, we have two different values for x and y,
depending on whether we go for the upper or lower bound.
However, we still cover the whole of the closed set [0;6],
which, of course, is in the same Borel class.

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

The Cool Giraffe wrote:
Stephen J. Herschkorn wrote/skrev/kaita/popisal/schreibt :
On Wed, 24 Jan 2007, The Cool Giraffe wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?

You need to find an example of two inequivalent metrics on some set.
Show that the open sets in each topology generate the same sigma-field.
For example, build the sigma-fields hierarchically. If at some point
you have the same collection of sets in each case, you are done.

first metric
min( sup( |x(t)-y(t)| ) ; 6) , both sup and min for t in [0;1]
second metric
min( inf( |x(t)-y(t)| ) ; 6) , both inf and min for t in [0;1]
Both x(t) and y(t) are continuous functions on [0;1].

The way i see it, we have two different values for x and y,
depending on whether we go for the upper or lower bound.
However, we still cover the whole of the closed set [0;6],
which, of course, is in the same Borel class.

Huh? Let I = the interval [0,1] and C(I) = the set of continuous
real-valued functions on I. You are trying to generate the same
sigma-field on C(I), not on I.

"The Cool Giraffe" <giraffen@viltersten.com> wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

Let your first metric, call it d1, be the usual metric on [0,1].

ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

Let your first metric, call it d1, be the usual metric on [0,1].

Thanks for the answer. However, i'd like to get a closer
clarification, since i'm very new to the subject.

When you say "usual metric" and then refer to the
topology created thereby, i understand that what
you mean is more or less the following.

d1(x,y) = |x-y| , x,y in [0;1]

and if for instance d1(x,y)=1/3 then the element in
the topology coresponding to it will be the set of all
such x's and y's that are exactly 1/3 from each other?
Or is it the intervals spanned between such x's and y's?

On Wed, 24 Jan 2007, The Cool Giraffe wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?

You can't. Consider the discrete metric and the usual metric for R.

"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

Let your first metric, call it d1, be the usual metric on [0,1].

Thanks for the answer. However, i'd like to get a closer
clarification, since i'm very new to the subject.

When you say "usual metric" and then refer to the
topology created thereby, i understand that what
you mean is more or less the following.

d1(x,y) = |x-y| , x,y in [0;1]

and if for instance d1(x,y)=1/3 then the element in
the topology coresponding to it will be the set of all
such x's and y's that are exactly 1/3 from each other?
Or is it the intervals spanned between such x's and y's?

The open sets of the d1 topology are unions of open (in [0,1])
intervals. For example, (0,.3) U (.4,, is an open subset of
[0,1], and so is [0,.3) U (.4,..

In contrast to this, the set [.5,1] is d2 open, but not d1 open.
(Indeed, [.5,1] = {x : d2(x,1) < .6}.)

"The Cool Giraffe" <giraffen@viltersten.com> wrote:

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

Let your first metric, call it d1, be the usual metric on [0,1].

Let your second metric, call it d2, be defined as follows:

d2(x,y) = |x-y| if 0 <= x < 1/2 and 0 <= y < 1/2
= |x-y|+1/2 if 0 <= x < 1/2 and 1/2 <= y <= 1
= |x-y|+1/2 if 1/2 <= x <= 1 and 0 <= y < 1/2
= |x-y| if 1/2 <= x <= 1 and 1/2 <= y <= 1.

With this metric [0,1] is homeomorphic to [0,1/2) U [1, 3/2) with the
usual topology.

The Borel sigma-algebra generated by the d2 topology is the same as
that generated by the d1 topology.

ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:

sniping and switching the order to avoid insanity

I mean that A is in the topology associated with D.
This is the case if and only if for each x in A there is
e(x)>0 such that tHe neighborhood
{y: D(x,y) < e(x) } is a subset of A.

Ah, now i see new things. In this case, i'd like to bother you
with the following two questions. :)

You said that:

The open sets of the d1 topology are unions of open (in [0,1])
intervals. For example, (0,.3) U (.4,, is an open subset of
[0,1], and so is [0,.3) U (.4,..

Is the above, namely that ]0,1/3[ as well as [0,1/3[ being
open subsets of [0,1] , true because of the fact that we can
unify our way down to zero as follows?

lim N->oo ( U( ]0+1/i;1/3[ , i=4 , N ) )

If not because of that, then why? I have this vague thought of
that since zero is on the edge of the original set then it needs
to be treated differently...

However, if it _IS_ because of that - what stops me from
going at it as follows,

lim N->oo ( U( ]1/2+1/i;3/4[ , i=7 , N ) )

hence creating a half-closed interval [1/2;3/4[ ?

Also, great thanks for the help this far. It's been of great
guidance and succour.

How can one prove that even inequivalent metrics can
result in the same class of borel sets?
Would it be sufficient to take two different metrics
both spanning over [0;1]?

"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:
ArtflDodgr wrote/skrev/kaita/popisal/schreibt :
"The Cool Giraffe" <giraffen@viltersten.com> wrote:

sniping and switching the order to avoid insanity

I mean that A is in the topology associated with D.
This is the case if and only if for each x in A there is
e(x)>0 such that tHe neighborhood
{y: D(x,y) < e(x) } is a subset of A.

Ah, now i see new things. In this case, i'd like to bother you
with the following two questions. :)

You said that:

The open sets of the d1 topology are unions of open (in [0,1])
intervals. For example, (0,.3) U (.4,, is an open subset of
[0,1], and so is [0,.3) U (.4,..

Is the above, namely that ]0,1/3[ as well as [0,1/3[ being
open subsets of [0,1] , true because of the fact that we can
unify our way down to zero as follows?

lim N->oo ( U( ]0+1/i;1/3[ , i=4 , N ) )

Yes, this representation of ]0,1/3[ shows that it is open.

lim N->oo ( U( ]1/2+1/i;3/4[ , i=7 , N ) )
hence creating a half-closed interval [1/2;3/4[ ?

But this union, namely
U{ ]1/2+1/i, 3/4[ , i=7,8,...},
is the open interval ]1/2,3/4[.