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 | | From: | William Elliot | | Subject: | Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 05:58:01 -0800 |
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 | Topological space S is homogeneous when for all x,y in S,
some auto-homeomorphism h:S -> S with h(x) = y.
Is a connected subspace of homogeneous space homogeneous?
No. [0,1] and [0,1) subset R are counterexamples.
Is an open subspace of homogeneous space homogeneous?
Is an open connected subspace of a homogeneous space homogeneous?
Counter examples, of course, are welcome.
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| | From: | Zbigniew Fiedorowicz | | Subject: | Re: Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 17:15:09 -0500 |
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| William Elliot wrote:
> Topological space S is homogeneous when for all x,y in S,
> some auto-homeomorphism h:S -> S with h(x) = y.
>
> Is a connected subspace of homogeneous space homogeneous?
> No. [0,1] and [0,1) subset R are counterexamples.
>
> Is an open subspace of homogeneous space homogeneous?
No. For example take the subset of the plane consisting of the disjoint
union of a disk and a punctured disk.
> Is an open connected subspace of a homogeneous space homogeneous?
This is tougher, for the reasons David Ullrich mentioned.
I believe the following would be a counterexample. Let
f: R \to S^1\times S^1
be given by
f(t) = (e^{2\pi i t}, e^{2\pi\alpha i t)
where \alpha is a fixed irrational. Note that f is injective.
Then f(R) (with the subspace topology coming from the torus)
is homogeneous.
Now let X be the complement in R of
[0, 1] \cup [2, 3]
Then f(X) is open in f(R) (since it is the complement of
the compact subset f([0,1])\cup f([2,3]). It is also connected
since f((-\infty,0)) is dense in f(R). However it seems to
me that f(X) is not homogeneous. I don't believe that there
is a auto-homeomorphism of f(X) sending f(1.5) to f(-1).
>
> Counter examples, of course, are welcome.
>
> ----
>
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| | From: | Zbigniew Fiedorowicz | | Subject: | Re: Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 17:25:31 -0500 |
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| Zbigniew Fiedorowicz wrote:
> Now let X be the complement in R of
> [0, 1] \cup [2, 3]
> Then f(X) is open in f(R) (since it is the complement of
> the compact subset f([0,1])\cup f([2,3]). It is also connected
> since f((-\infty,0)) is dense in f(R). However it seems to
> me that f(X) is not homogeneous. I don't believe that there
> is a auto-homeomorphism of f(X) sending f(1.5) to f(-1).
>
Here's an argument for the last assertion. Using covering space
theory, we can see that f(X) has three path components:
f((-\infty,0)), f((1,2)) and f((2,\infty)). Any auto-homeomorphism
of f(X) would have to permute them. But f((-\infty,0)) is dense
in f(X), whereas f((1,2)) is not. QED.
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| | From: | William Elliot | | Subject: | Re: Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 23:10:51 -0800 |
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| From: Zbigniew Fiedorowicz
Newsgroups: sci.math
Subject: Re: Homeogeneous Spaces
> William Elliot wrote:
> > Topological space S is homogeneous when for all x,y in S,
> > some auto-homeomorphism h:S -> S with h(x) = y.
>
> > Is a connected subspace of homogeneous space homogeneous?
> > No. [0,1] and [0,1) subset R are counterexamples.
>
> > Is an open subspace of homogeneous space homogeneous?
> No. For example take the subset of the plane consisting
> of the disjoint union of a disk and a punctured disk.
Quick and simple.
>> Is an open connected subspace of a homogeneous space homogeneous?
> This is tougher, for the reasons David Ullrich mentioned.
> I believe the following would be a counterexample.
> Let f:R to S^1 times S^1 be given by
> f(t) = (e^{2pi.it}, e^{2pi.alpha.it})
> where alpha is a fixed irrational. Note that f is injective.
> Then f(R) (with the subspace topology coming from the torus)
> is homogeneous.
An almost periodic wrapping of a line around a torus.
> Now let X be the complement in R of [0, 1] cup [2, 3]
> Then f(X) is open in f(R) (since it is the complement
> of the compact subset f([0,1]) cup f([2,3]).
> It is also connected since f((-infty,0)) is dense in f(R).
Yes, f((-oo,0)), f((1,2)) and f((3,oo) are connected.
Density doesn't imply connected. That f(X) is connected must
have to do with density, wrapping around and path components.
> However it seems to me that f(X) is not homogeneous.
> I don't believe that there is a auto-homeomorphism
> of f(X) sending f(1.5) to f(-1).
> Here's an argument for the last assertion. Using covering
> space theory, we can see that f(X) has three path components:
> f((-infty,0)), f((1,2)) and f((2,infty)). Any auto-homeomorphism
> of f(X) would have to permute them. But f((-infty,0)) is dense
> in f(X), whereas f((1,2)) is not. QED.
Ya, f((1,2)) is a locally connected wrap once around
the tube. The other two aren't locally connected.
Couldn't we conclude, f(X) isn't homogeneous for having different
path components, one locally connected, two not locally connected?
Well done. This inquiry was motivated by considering locally connected,
locally homogeneous spaces. Does one of those always have a base of open
connected homogeneous sets?
But as you shown, homogeneous, even for open connected subsets doesn't
inherite, thus demolishing a quick application of an easy theorem about
merging base properties. The other way, is an homogenous subset of a
connected space connected, is of course, fanstasy.
If C is a base of connected sets and H is a base of homogeneous sets,
and further if U in C, V in H, U subset V, is U homogeneous?
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| | From: | sim_stef at yahoo.com | | Subject: | Re: Homeogeneous Spaces | | Date: | 24 Jan 2005 01:00:56 -0800 |
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| Zbigniew Fiedorowicz wrote:
[.....]
> Now let X be the complement in R of
> [0, 1] \cup [2, 3]
> Then f(X) is open in f(R) (since it is the complement of
> the compact subset f([0,1])\cup f([2,3]). It is also connected
> since f((-\infty,0)) is dense in f(R). However it seems to
> me that f(X) is not homogeneous. I don't believe that there
> is a auto-homeomorphism of f(X) sending f(1.5) to f(-1).
Sure, f(X) has two arc components, which are not homeomorphic - one of
them is locally connected, the other one is not. (The points f(1.5) and
f(-1) lie in different arc components.)
Your example may be modified in order to obtain a compact example.
Let h be a minimal self-homeomorphism of the Cantor set K and T(h) be
the mapping torus of h. Then taking two points x,y in T(h) from one and
the same arc component, T(h)\{x,y} is open and connected, but
non-homogeneous for similar reasons.
Of course, this example is not locally connected. It is interesting to
find the "best" possible example, say - compact metric, connected
and locally connected,... is that too much?
Simeon
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| | From: | David C. Ullrich | | Subject: | Re: Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 14:32:31 -0600 |
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| On Sun, 23 Jan 2005 05:58:01 -0800, William Elliot
wrote:
>Topological space S is homogeneous when for all x,y in S,
> some auto-homeomorphism h:S -> S with h(x) = y.
>
>Is a connected subspace of homogeneous space homogeneous?
> No. [0,1] and [0,1) subset R are counterexamples.
>
>Is an open subspace of homogeneous space homogeneous?
>Is an open connected subspace of a homogeneous space homogeneous?
Surely not - I mean there's simply no reason why one would
imagine this is so.
Ok, that's not a counterexample, quite. Regarding your
search for counterexamples, note that any open connected
subset of R^n _is_ homogeneous. (Pf: If O is such a set
and a, b in O then there exists a set B homeomorphic to
a closed ball containing a and b in the interior and
contained in O. Any two points of a ball are swapped
by some homeomorphism of the ball fixing points of the
boundary - that homeomorphism hence extends to a
homeomorphism of O.)
>Counter examples, of course, are welcome.
>
>----
************************
David C. Ullrich
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| | From: | William Elliot | | Subject: | Re: Homeogeneous Spaces | | Date: | Sun, 23 Jan 2005 23:09:52 -0800 |
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| From: David C. Ullrich
Newsgroups: sci.math
Subject: Re: Homeogeneous Spaces
> William Elliot wrote:
>> Topological space S is homogeneous when for all x,y in S,
>> some auto-homeomorphism h:S -> S with h(x) = y.
>
>> Is a connected subspace of homogeneous space homogeneous?
>> No. [0,1] and [0,1) subset R are counterexamples.
>
>> Is an open subspace of homogeneous space homogeneous?
>> Is an open connected subspace of a homogeneous space homogeneous?
> Surely not - I mean there's simply no reason why one would
> imagine this is so.
Because then one would have a base of open connected homogeneous
sets any for locally connected and locally homogeneous space.
> any open connected subset of R^n _is_ homogeneous.
For R this is immediate, because an open connected subset is
open interval, hence homeomorphic to R, which is homogeneous.
> (Pf: If O is such a set and a, b in O then there exists a set B
> homeomorphic to a closed ball containing a and b in the interior
> and contained in O. Any two points of a ball are swapped by some
> homeomorphism of the ball fixing points of the boundary - that
> homeomorphism hence extends to a homeomorphism of O.)
Ok, I suppose. Seems alright.
Aren't open connected subsets of R^n homeomorphic to an open ball?
No, they have to be simply connected. Then that is so.
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