Re: Topology question

On Nov 26, 12:31 pm, Zbigniew Karno <Zbigniew.Ka...@xxxxxxxxx> wrote:
On 26 Lis, 06:57, omega <blue_jag...@xxxxxxxxx> wrote:





Zbigniew Karno wrote:
On 25 Lis, 18:02, omega <blue_jag...@xxxxxxxxx>
wrote:
Let X be a locally compact, complete metric space.
Suppose that X is path-connected,
and that the covering dimension dim X = +oo. Is it
true that X contains a copy of
its own Alexandroff compactification X* ?

No.

(Is the space X non-compact?)

--
Z. Karno

Thx. A little counterexample would be helpful.

For each integer n >= 1 in R^n (with the usual topology) take the
closed ball B_n = {x in R^n : ||x|| =< 1/n}. Note that B_n is a
compact space with the diameter =< 1/n and dim B_n = n. The spaces
B_n can be considered as subspaces of the Hilbert space H via the
embedding
R_n >--> H, (x_1,x_2,...,x_n) |--> (x_1,x_2,...,x_n,0,0,...)

Now in the space H x R (with the product topology) consider the
subspace

X = {0} x [0,+\infty) \cup \bigcup {B_n x {n} : n=1,2...,+\infty}

X looks like as the infinite ray, with infinite collection of
balls attached at positive integers, intersecting each at its
center. Since the sequence of the diameters of B_n tending to 0,
it follows that X is complete, and also path-connected and locally
compact, but non-compact. So, X* exists and it is clear that X
does not contain none topological copy of X*. Obviously, dim X =
\infty (because dim B_n = n and n = 1,2,...).

Also, could you provide me with a sufficient condition for a locally compact (and noncompact, of course), path-connected, complete metric space X, in order that X contains a copy of X*?

For instance (maybe the following is true):
If X is a complete, path-connected, locally compact and noncompact
metric
space with the property that
(*) for every compact subset K of X there exists an open subset
V of X with compact closure, containing K and also there exists
an embedding h : X --> V such that h(x) = x for all x in K,
then X contains a homeomorphic copy of X*

Regards, Z. Karno- Hide quoted text -

- Show quoted text -

Maybe X=R is a counterexample to your statement
.

Relevant Pages

  • Re: Topology question
    ... Now in the space H x R (with the product topology) consider the ... If X is a complete, path-connected, locally compact and noncompact ... for every compact subset K of X there exists an open subset ...
    (sci.math)
  • Re: seperated-proper-complete
    ... With the Zariski topology, isn't every affine space compact? ... Certainly any affine variety (or ...
    (sci.math)
  • Re: quasi-compact topological vector spaces
    ... Do you distinguish closure of ball, cl Band closed ball ... > literature dealing with the topology on Banach spaces defined below. ... You appear to be considering a variant of cocompact or compact complement ...
    (sci.math.research)
  • Re: Why Compact open topology?
    ... The compact-open topology is not the natural one. ... homis Top(i.e. the whole set of continuous maps from X to ... colimits of compact Hausdorff spaces. ...
    (sci.math)
  • Re: Topology with first countable.
    ... Let S have the cofinite or the cocountable topology. ... What I meant here was that "the complement of A is compact". ... Another reason I didn't guess that is that it's locally compact ...
    (sci.math)
Loading