On Mon, 06 Oct 2008 15:23:38 -0400, quasi <qu...@null.set> wrote: >If U is a nonempty, non-dense open subset of R^n, with n > 1, must the >boundary of U contain a nontrivial path?
If the answer to my question above is "yes", then perhaps we can try for more:
If U is a nonempty, non-dense open subset of R^n, with n > 1, must the boundary of U contain a subset which is homeomorphic to R^(n-1)?
>>> If U is a nonempty, non-dense open subset of R^n, with n > 1, must the >>> boundary of U contain a nontrivial path? >> If the answer to my question above is "yes", then perhaps we can try >> for more:
>> If U is a nonempty, non-dense open subset of R^n, with n > 1, must the >> boundary of U contain a subset which is homeomorphic to R^(n-1)?
> No. Any open half plane.
Are you saying that the boundary of an open half plane (in R^2, I suppose) does not contain a subset which is homeomorphic to R? But the boundary itself *is* homeomorphic to R!
> No, even if the open set is bounded.
> The set inclosed by the topologist's sin curve nipping it's tail.
And you claim that "the topologist's sin curve nipping it's tail" has no subset homeomorphic to R?
<ma...@hevanet.remove.com> wrote: >On Mon, 6 Oct 2008, quasi wrote:
>> On Mon, 06 Oct 2008 15:23:38 -0400, quasi <qu...@null.set> wrote:
>> >If U is a nonempty, non-dense open subset of R^n, with n > 1, must the >> >boundary of U contain a nontrivial path?
>> If the answer to my question above is "yes", then perhaps we can try >> for more:
>> If U is a nonempty, non-dense open subset of R^n, with n > 1, must the >> boundary of U contain a subset which is homeomorphic to R^(n-1)?
>No. Any open half plane.
What are you talking about?
The boundary of a half plane in R^2 is a line, hence is homeomorphic to R^1.
>No, even if the open set is bounded.
>The set inclosed by the topologist's sin curve nipping it's tail.
I think you misread the question.
Although I'm not sure precisely which open set you are suggesting, let me point out that topologist's sine curve, with or without its tail, certainly has a _subset_ which is homeomorphic to R^1.
On Tue, 7 Oct 2008, quasi wrote: > <ma...@hevanet.remove.com> wrote: > >On Mon, 6 Oct 2008, quasi wrote: > >> On Mon, 06 Oct 2008 15:23:38 -0400, quasi <qu...@null.set> wrote:
> >> >If U is a nonempty, non-dense open subset of R^n, with n > 1, must the > >> >boundary of U contain a nontrivial path?
> >> If the answer to my question above is "yes", then perhaps we can try > >> for more:
> >> If U is a nonempty, non-dense open subset of R^n, with n > 1, must the > >> boundary of U contain a subset which is homeomorphic to R^(n-1)?
> >No. Any open half plane.
> >No, even if the open set is bounded.
> >The set inclosed by the topologist's sin curve nipping it's tail.
William Elliot <ma...@hevanet.remove.com> wrote: > On Mon, 6 Oct 2008, quasi wrote: > > On Mon, 06 Oct 2008 15:23:38 -0400, quasi <qu...@null.set> wrote:
> > >If U is a nonempty, non-dense open subset of R^n, with n > 1, must the > > >boundary of U contain a nontrivial path?
> > If the answer to my question above is "yes", then perhaps we can try > > for more:
> > If U is a nonempty, non-dense open subset of R^n, with n > 1, must the > > boundary of U contain a subset which is homeomorphic to R^(n-1)?
> What if the boundary is a fractile? > What if the open set is the interior of the Snowflake curve?
> What if the boundary is a loop of Brownian motion? > Hm, what would the interior look like?
These are not counterexamples (where "path" means "homeomorphic image of [0,1]") but I expect there are counterexamples. Maybe in Steen & Seebach...
