New answers tagged gn.general-topology
13
votes
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Mistake on article about Bohr compactification?
The problem is in the proof of Theorem 2. We have two maps to begin with: the map $b:\mathbb{R}\to b\mathbb{R}$ of the Bohr compactification (called $\tau$ in the paper), and an embedding $e$ of $\...
- 11.4k
6
votes
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Codimension zero embeddings and maps with small fibers
Isn't the following an example?
Build $M$ starting from the plane by attaching an infinite sequence of handles whose size decreases to zero.
Take for $N$ the disjoint union of $M_k$, where $M_k$ is ...
- 2,772
12
votes
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Is $X\times X$ homeomorphic to $X$ for a space of probability measures?
The answer is yes, following from Klee's extension of Keller's theorem on homeomorphisms of infinite dimensional compact convex sets:
Klee, V. L., Some topological properties of convex sets, Trans. Am....
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9
votes
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Lower bound on dimension required to disconnect manifold?
Let my write up my comment as an answer. The easiest argument I know uses sheaf cohomology. It requires working through a bit of theory, but ultimately gives a very flexible tool for proving all sorts ...
- 28.7k
4
votes
Jordan plane curve such that $\frac{d(g(x),g(y))}{d(x,y)}\to0$?
Sure. It is a standard fact that the von Koch snowflake $C$ can be parametrized by
$$ f\colon S^1 \rightarrow C$$
where
$$ C^{-1}|x-y|^{1/p} \leq |f(x)-f(y)| \leq C|x-y|^{1/p},$$
$p\in (1,2)$ is the ...
- 41
8
votes
Accepted
Fundamental group of the grid on $\mathbb{R}^\mathbb{N}$
Let $G_n$ be your "grid" in $\mathbb{R}^n$ for $n\geq 2$. Let $p_{n,n-1}:G_n\to G_{n-1}$ be the usual projection map. Then $X$ can be identified canonically with the inverse limit $\...
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7
votes
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Connected open sets in the topology generated by the collection of connected open sets
Take $\mathbb{Q}$ and add two points at infinity, $\infty_1$ and $\infty_2$, topologizing so that $\mathbb{Q} \cup \{\infty_i\}$ is the only open proper subset containing $\infty_i$ for $i \in \{1,2\}$...
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3
votes
Preimage of a sublocale by a morphism of locales: description by nucleus?
Here is I think a counter-example to the precise proposed formula in the question.
Take $X= \mathbb{Q}$ with the discrete topology, with $Y = \mathbb{R}$ withe its usual topoly and the map $f:X \to Y$ ...
- 42.4k
10
votes
Accepted
Are there any tests for knowing whether a topological space admits a CW structure?
Every compact topological manifold $M$ has the homotopy type of CW-complex. So, in the compact case, there is no such invariant of the type you are looking for (or, at least, such an invariant cannot ...
- 66.3k
4
votes
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Even covers and collectionwise normal spaces
As implied by Tyrone, the claim that a fully normal space is strongly collectionwise normal is true. See, for example, the result in H. J. Cohen's paper Sur un problème de M. Dieudonné (translated ...
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4
votes
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Is the interval topology on ${\cal P}(\omega)/(\text{fin})$ connected?
The space is hyperconnected and thus also connected. Indeed, it suffices to show that, for any $x_1, \cdots, x_n \in \mathcal{P}(\omega)/(\text{fin})$ which are not $[\omega]$, and for any $y_1, \...
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6
votes
Is there an 'unnatural' topological construction of an algebraically closed field of positive characteristic?
This is not directly an answer to any of your questions as stated but a riff on the theme of "what does $\overline{\mathbb{F}_p}$ look like?" The best answer to this question I've found so ...
- 118k
2
votes
Accepted
A question about G-Hewitt spaces
The statement is analogous to the general result that, for Tychonoff spaces, compactness is equivalent to pseudo-compactness plus realcompactness, see the beginning of section 3.11 in Engelking's ...
- 11.4k
3
votes
Cohomology version of Moore space
As Allen Hatcher answered, there is no space whose cohomology is a countable direct sum of $\mathbb{Z}$'s in a single degree, and the cohomology of a wedge of spheres is instead a product.
However, ...
- 10.8k
8
votes
Cohomology version of Moore space
One needs to distinguish between the direct sum and the direct product of a collection of groups. For a countably infinite collection of copies of $\mathbb Z$ the direct sum of these groups is a free ...
- 20k
0
votes
Continuum-distanced complete, ultrametric space
Let's put put the question in a bit more general context: Which type of metric spaces admit the property that nested closed balls can have empty intersection:
(1) Complete: Yes (as in the original ...
0
votes
Continuum-distanced complete, ultrametric space
Let
$$ X\,\ :=\,\ (1;2)\ =\ \{x\in\mathbb R: 1< x< 2\} $$
The distance in $\ X\ $ is defined by:
$$ \forall_{x\ y\in X}\quad d(x\ y)\ :=\ \max(x\ y) $$
This ultra-metric space is discrete hence ...
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3
votes
Accepted
Relation between $\mathbb{R}$ and the metric space of bounded functions $f:\mathbb{N}\to\mathbb{N}$
For every $K$ the subspace $X_K$ of all functions bounded by $K$ is homeomorphic to the Cantor set, as $X_K=\{1,2,\ldots,K\}^\mathbb{N}$ and the metric induces the product topology.
It follows that $\...
- 11.4k
9
votes
Continuum-distanced complete, ultrametric space
This is impossible, because
for any complete normed space $X$ and any nonincreasing sequence $(B_n)$ of nonempty closed balls in $X$ we have $\bigcap_n B_n\ne\emptyset$.
Indeed, take any ...
- 128k
4
votes
Accepted
Strong ultralimits?
The first thing to to with a definition like that is test it against some familiar examples. The sequence $\langle 2^{-n} : n\in\mathbb{N}\rangle$ converges to $0$ in $\mathbb{R}$ and it would seem ...
- 11.4k
1
vote
Homeomorphism and boundary of a complementary component
The answer to Question 2 is negative: Let $Y$ be a continuum consisting of the segment $[-1,1]\times\{0\}$ to which a sequence of half circles $C_n$ with radius $\frac1n$ lying in the upper halfplane ...
- 770
1
vote
Homeomorphism and boundary of a complementary component
For the modified question, here is a counter-example:
First, note that the Cantor set $K$ is (topologically) homogeneous: the group of homeomorphisms acts transitively. (One way to see this is by ...
- 12.2k
6
votes
For $\mathbb R^n \times Q \cong \mathbb R^m \times Q $ must $n = m$? ($Q$ is the Hilbert cube)
Note that $\mathbb{R}^n\times Q$ is proper homotopy equivalent to $\mathbb{R}^n$.
If they were homeomorphic, we would obtain a proper homotopy equivalence between $\mathbb{R}^m$ and $\mathbb{R}^n$.
...
- 11.1k
2
votes
Accepted
Is every subgroup closed in this complete, nondiscrete topological group?
The metric induces the product topology, so the group $G$ is compact. The direct sum of $\mathbb{Z}$ many copies of $G'$ is a countable dense subgroup, but not the whole group, so it is not closed.
- 11.4k
2
votes
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Compact Hausdorff spaces as a cocompletion of profinite sets
The category of compact Hausdorff spaces is the pretopos completion of the category of profinite sets. It means that it is obtained by adding freely all quotients by equivalence relations, which are ...
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4
votes
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Topological property of the space of probability measures
I'm not sure whether or not these subsets are homeomorphic, but there can not be such a map $\theta$ because $P_s$ is a $G_{\delta}$ set. Compare Theorem 1.2 here.
If we had a $\theta$ as desired, ...
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