Download Jordan Brower

Jordan-Brouwer Separation Theorem and Invariance of
Domain
#1
Anonymous217
Posts: 344
So I'm having trouble
understanding how
these two are related,
i.e., how one proves
the other.
I understand the ideas
behind both of them:
For J-B, you're
basically taking R^n
and throwing in a
sphere, so the inside of
the sphere is bounded
and everything outside
the sphere is
unbounded. For
Invariance of
Domain, it's pretty
obvious just by the
definition (the image of
an open subset of R^n
is open).
However, I don't really
see a relationship
between the two. Can
anyone give some
insight?
Also, I was curious why
we only need injectivity
for both of them, where
surjectivity is basically
unnecessary in any
possible proof. What
makes 1-1 important?
Anonymous217
View Public Profile
Find More Posts by Anonymous217
science news on PhysOrg.com
PhysOrg.com
>> Intel takes stake in ASML
>> Polymer power: Triboelectric generator produces
electricity by harnessing friction between surfaces
>> Marcellus brine migration likely natural, not manmade: study
Sep2-11, 08:02 AM
Re: JordanBrouwer
Separation
Theorem
and
Invariance
of Domain
#2
quasar987
Posts: 4,656
Recognitions:
PF Contributor
Homework Helper
Science Advisor
I was
unaware
that these
theorems
were
equivalent
. Where
did you
hear such
a claim?
The only
way I
know
them to be
related is
that they
are both
obviouslooking
statement
that no
one has
been able
to prove
for a long
time. That
is, until
homology
came
along.
quasar987
View Public Profile
Find More Posts by quasar987
Sep2-11, 10:19 AM
mathwonk
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#3
(zeroth) homology measures components of a space hence
can be used to prove the separation thm. components of an
open set in Euclidean space are open, so knowing about
components can be used to prove inv of domain.
If S is an n-1 sphere and B an n ball, and f an injective
continuous map, separation implies the complement of the
image of f(S) in R^n has 2 components and the complement
of f(B) has only one.
Posts: 8,841
Recognitions:
since f(B-S) is connected, it is thus a component of the open
complement of f(S), hence f(B-S) is open.
Homework Helper
Science Advisor
Since every open set is a union of such balls B, this shows
that Jordan separation implies invariance of domain.
mathwonk
View Public Profile
Find More Posts by mathwonk
Sep2-11, 10:27 AM
mathwonk
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#4
as a weak version of the other direction, suppose f is an
injective continuous map from B to R^n, and we assume
f(B) has connected complement. Then if we knew that f(B-S)
were open, it would follow, since f(B-S) is also connected,
that the complement of f(S) has 2 components. I.e.
invariance of domain does imply that the complement of f(S)
has one more component than the complement of f(B).
Posts: 8,841
Recognitions:
Homework Helper
Science Advisor
mathwonk
View Public Profile
Find More Posts by
mathwonk
Sep2-11, 03:28 PM
Anonymous217
Posts: 344
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#5
^^ This is the type of answer I was looking for. It makes
complete sense now; thanks.
Anonymous217
View Public Profile
Find More Posts by Anonymous217
Sep2-11, 09:17 PM
mathwonk
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#6
notice the weak direction is weak because it is not clear,
indeed probably not true, that an injection from S^n-1 to
R^n extends to an injection on B^n.
Posts: 8,841
Recognitions:
Homework Helper
Science Advisor
mathwonk
View Public Profile
Find More Posts by mathwonk
Sep2-11, 09:33 PM
mathwonk
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#7
you are welcome.
Posts: 8,841
Recognitions:
Homework Helper
Science Advisor
mathwonk
View Public Profile
Find More Posts by
mathwonk
Sep2-11, 11:59 PM
Re: Jordan-Brouwer Separation Theorem and Invariance of
Domain
#8
Bacle
Mathwonk:
Posts: 662
I think we may be able to use Tietze's theorem that a
continuous, real-valued map from a closed subspace of a
normal space (I think the ball B^n, as a metric space, is
normal, and its boundary S^(n-1) is closed in B^n--the
interior seems to be open) X, extends into the whole space,
tho I am not sure if there is a version for maps into R^n;
maybe we can argue component-wise to get a continuous
map.
From: "Daniel Giaimo" <[email protected]>
Subject: Re: Homeomorphisms of the sphere
Date: Sat, 4 Dec 1999 23:03:54 -0800
Newsgroups: sci.math
Keywords: Invariance of Domain theorems
Romain Brette <[email protected]> wrote in message
news:[email protected]...
>Le dim, 05 déc 1999, vous avez écrit :
>>Romain Brette <[email protected]> wrote in message
>>news:[email protected]...
>>> Does anybody know how to prove that the n-dimensional sphere is not
>>> homeomorphic to a (strict) part of itself ? It's quite easy for the
circle,
>>> but turns out to be quite hard in general.
>>
>>
Suppose S^n is homeomorphic to a strict subset of itself.
>>
Let f:S^n->f(S^n) be such a homeomorphism. Then f(S^n) != S^n,
>>therefore there exists a point x_0 in S^n such that x_0 is not in
f(S^n).
>>Therefore f actually maps into S^n\{x_0} which is homeomorphic to
R^n. Now,
>>by Invariance of Domain, f(S^n) is open in R^n as f is 1-1 and
continuous.
>
> I guess you mean the projection of f(S^n) in R^n. But what is
"Invariance of
>Domain" ? I can't see why this should be open. The thing is if A is
open and f
>is a 1-1 and continuous, then f(A) is open _relatively_ to the range,
which is
>f(S^n) (which means, it's the intersection of an open set with
f(S^n)). So it
>doesn't tell anything about the image of the domain. For example, take
a
>homeomorphism from the disc to a part of it (a contraction)
>Then the image of the domain is not open. Can you state what you mean
by
>"Invariance of Domain" ?
Invariance of Domain is a theorem which states that if f:M->N is a
map
between n-dimensional manifolds without boundary which is 1-1 and
continuous, then f is an open map. In fact, you really only need the
weaker
form which says that if f:S^n->S^n is open and continuous then it is
open.
---Daniel Giaimo
Remove nospam. from my address to e-mail me. |
dgiaimo@(nospam.)ix.netcom.com
^^^^^^^^^<(Remove)
|--------BEGIN GEEK CODE BLOCK--------|
anyway.
|Version: 3.1
|
|GM d-() s+:+++ a--- C++ UIA P+>++++ |
|L E--- W+ N++ o? K w>--- !O M-- V-- |
|PS? PE? Y PGP- t+(*) 5 X+ R- tv+(-) |
|b+@ DI++++ D--- G e(*)>++++ h->++ !r |
|!y->+++
|
|---------END GEEK CODE BLOCK---------|
Ros: I don't believe in it
Guil: What?
Ros: England.
Guil: Just a conspiracy of
cartographers, you mean?
=======================================================================
=======
From: "Daniel Giaimo" <[email protected]>
Subject: Re: Homeomorphisms of the sphere
Date: Sun, 5 Dec 1999 12:19:16 -0800
Newsgroups: sci.math
Edward C. Hook <[email protected]> wrote in message
news:[email protected]...
[deletia; similar to above --djr]
>|> by Invariance of Domain, f(S^n) is open in R^n as f is 1-1 and
continuous.
>
>
How do you derive this from Invariance of Domain ?? AFAICR, that
>
theorem asserts that either every embedding of a space X into R^n
>
has an open image or none of them do.
There are several theorems known as Invariance of Domain. The one
I am
using is the theorem that if M and N are topological n-manifolds
without
boundary, and f:M->N is 1-1 and continuous, then f is open.
---Daniel Giaimo
[sig deleted; see above --djr]
=======================================================================
=======
From: [email protected] (Lee Rudolph)
Subject: Re: Topology question in R^n
Date: 14 Jun 1999 11:35:18 -0400
Newsgroups: sci.math
>>
>>
>>
>>
Let A and B be two subsets of R^n. Let
f: A -> B be a homeomorphism.
If A is an open set (for the topology induced by R^n),
is B always open?
[deletia --djr]
In fact, the question as understood by me (and, I believe, as meant by
the questioner) is not trivial (at least not for n>1); that the answer
is "yes" (Brouwer's theorem on "invariance of domain") for all n
depends
on properties of R^n that are not shared by all spaces.
>A homeomorphism is a bijective continuous map whose inverse is also
>continuous, so it maps open sets to open sets and closed sets to
closed
>sets by definition of the continuity of the inverse map.
Consider the space X = the union in R^2 of the x-axis and the y-axis,
with the topology induced by R^2. Let A be the (strictly) positive
x-axis. Then A is an open subset of the space X. Let B be the
entire x-axis. Then B is not an open subset of the space X.
Yet A and B (with their topologies induced by X, which are
of course their topologies induced by R^2) are homeomorphic.
Notice that X does not share various salient properties with R^1.
Notably, the local homology of X at the point (0,0) is different
from its local homology at other points. This is relevant.
Lee Rudolph