New answers tagged gn.generaltopology
7
The answer is clearly no, and $S^1$ (or, in general, $S^n$) provides an easy counterexample.
The topological spaces not admitting a homeomorphism onto a proper subset are called topologically finite, and were studied in
I. Tsereteli, Topological finiteness, Bernstein sets, and topological rigidity, Topology Appl. 159, No. 6, 16451653 (2012). ZBL1241....
2
The answer is NO, i.e. the following nonsingleton connected subspace
$\ X\ $ of Euclidean $\ \mathbb R^2\ $ does not have any proper subspace
homeomorphic to $X$:
$\qquad X\ :=\ S\,\cup\,\{(1\ 0)\ \ (1\ 0)\} $
where
$\qquad S\ :=\ \{(x\,\ \sin\frac 1{1x^2})\,:\ 1<x<1\} $

PROOF: Let $\ X'\subseteq X\ $ be homeomorphic to $\ X.\ $
The every ...
0
Welp, this wasn't my finest moment: unless I'm missing something, we can translate everything much more easily than I thought at first. (I may indeed be missing something, however, and I'll wait a while before accepting this just in case.)
For simplicity I'll look at bases only; it won't make a serious difference.
Suppose I have enumerated bases $\mathcal{...
15
The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $\mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $\mathbb R$ endowed with such topology $\tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the ...
1
For boundedness of sets the statement is false. The Wikipedia quote is for linear operators.
A counterexample for sets: $X=\mathbb{R}^\omega$ in the product topology is a metric locally convex TVS. No neighbourhood of $0$ (like the open balls which are $d$bounded) can be "absorbingbounded" (Because it contains a product basic open set which has almost ...
7
There exists a metrizable topological group $H$ such that $H \setminus \{e\}$ is rigid (see Theorem 6.1 in van Mill´s paper: A topological group having no homeomorphisms other than translations).
Exercise: Without knowing anything else about $H$, show that $X=H \setminus\{e\}$ satisfies condition 2 in the OP.
6
For $n\in\mathbb N$ let $U_n=\{m\in\mathbb N:m\geq n\}$. Then $\tau=\{\varnothing\}\cup\{U_n:n\in\mathbb N\}$ is a topology on $\mathbb N$. This space is rigid because $n$ is characterized as the unique element contained in exactly $n+1$ open sets. However, for any $n,m$ the sets $U_n,U_m$ are homeomorphic through a map $k\mapsto kn+m$ which takes $n$ to $m$...
1
As Taras Banakh says, it really depends on the underlying group. Some comments in the direction of having a unique CH group topology (which of course is not the case in general):
Profinite groups are residually finite, whereas connected compact Hausdorff groups are divisible. So if a group $G$ admits a CH group topology, then $G$ has a largest divisible ...
2
If $\mathcal{X} = (\prod_{i=1}^\infty X_i ) \times X$ is endowed with the product topology, then to show that $F = (\prod_{i=1}^N S_i) \times f$ is topologically transitive (that is, given any pair of open subsets $U, V \subset \mathcal{X}$, there exists $n\in \mathbf{N}$ such that $F^n(U) \cap V \ne \emptyset$), it suffices to consider open subsets of the ...
4
The answer is yes for countable graphs:
Fix an infinite graph $G$ and a bijective homomorphism $f:G \to G$. Define $c:[G]^2 \to 2$ as $c(\alpha,\beta)=1$ if $\{f\alpha, f\beta\} \in E(G)$ and $c(\alpha,\beta)=0$ otherwise. Since $G$ is infinite, by Ramsey´s Theorem there is an infinite $c$homogeneous $H \subseteq G$. If $c \upharpoonright [H]^2$ is ...
1
There is nothing to say in the general case, but in the case of a continuous dcpo $D$ there is a wellknown construction of certain bases for the Scott topology on $D$.
We say $d$ is way below $e$, or $d \ll e$ if for each directed set $(d_i)_{i \in I}$ such that $e \leq \bigvee_{i \in I}d_i$, there exists $i \in I$ such that $d \leq d_i$. For example, if $...
5
The number of compact Hausdorff group topologies on a given group strongly depends on the algebraic structure of the group.
For example, any finitedimensional torus $(\mathbb R/\mathbb Z)^n$ has $2^{\mathfrak c}$ automorphisms, among which only finitely many continuous. This implies that $(\mathbb R/\mathbb Z)^n$ has $2^{\mathfrak c}$ pairwise incomparable ...
1
Here I am gathering information in the comments along with some information of my own to form an answer.
A $P$space is a regular space where every countable intersection of open sets is open. There are many examples of $P$spaces which are not discrete. For example, the cocountable topology is a topological space $(X,\mathcal{T})$ where $X$ is an ...
1
I'm always afraid of confusing lsc and usc, but what about $d=1$ and $U=\mathbb R\setminus\mathbb Q$ and $f(0)=0$ and $f(x) =1$ for $x\neq 0$?
3
I think it is possible to construct a totally disconnected compact $K\subset \mathbb R^2$ such that $f:\overline{M}\to\mathbb R^2$ would send at least two points to the origin. I will make a certain assumption about a polygonal neighborhood around a polygonal path. I can try to fill this step in or find a reference for that step if you aren't convinced by it....
4
The answer is no. Consider poset consisting of infinitely many incomparable elements $a_1,a_2,\dots$ and a single element $b$ larger than them all. Then $A=\{a_1,a_2,\dots\}$ is closed in the Scott topology (note it has no directed subsets with more than one element).
On the other hand, consider the topology generated by Scottclosed ideals. If an ideal ...
4
[Edit] I think the following argument, made possible by the help of Anton Petrunin in the comments (thank you very much!), does the trick. I left the original sketch below for historical/affective reasons.
Suppose that $K\subset\mathbb R^d$ is a compact satisfying the above nondistortion condition. Then the argument of Gromov described above shows that ...
4
This is an answer you might not accept.
Looking carefully at the book, I saw that the title of the chapter is "Riemannian manifolds with boundary and subsets of $\mathbb{R}^n$ with smooth boundary".
In a compact manifold, you can always choose such a length minimizing curve (the rough argument is that the unit tangent bundle is compact, you should maybe be ...
3
Your question is related to a pair of old questions of A.V.Arhangel'skii. First of all note that in a regular space, for every point $x$ and every $G_\delta$ set G containing $x$ there is a closed $G_\delta$ $H$ contained in $G$ such that $x \in H$. So the topology generated by the closed $G_\delta$ sets of a compact Hausdorff space $X$ coincides with the "$...
0
I think that maybe I should have thought about this a little harder before posting, because it seems like there is a fairly easy positive answer:
Let $X=[0,1]$ with a standard enumeration of the rational numbers. Let $f:[0,1]_c \rightarrow \mathbb{R}$ be a computable function that is unbounded (an easy example is to let $f(x)=\frac{1}{g(x)}$ where $g:[0,1]\...
4
$2^\kappa$ for uncountable $\kappa$ is a ZFC example of an almost sequential nonsequential space.
The easiest way to see why it's not sequential is to note that there is a set $A \subset 2^\kappa$ and a point $p \in \overline{A}$ such that $p$ is not in the closure of any countable subset of $A$. It suffices to take $A=\{x \in 2^\kappa: x^{1}(1) < \...
3
No.
Just looking at countably many generators we can produce a continuum of pairwise disjoint clopen subsets of $X$. Moreover, since $A=2^{\aleph_0}$, we have that $2^{\aleph_0} \leq c(X) \leq d(X) \leq w(X) \leq 2^{\aleph_0}$, where $c$, $d$ and $w$ denote cellularity, density and weight respectively.
1
There is a new and attractive book on Index theory and applications to Physics by Booss and Bleecker which covers all the necessary analysis background. To quote from its preface :
In order to enjoy reading or even work through Parts IIII, we expect the readerto be familiar with the concept of a smooth function and a complex ...
8
Note that replacing "wellordered" by "linearlyordered" produces an equivalent property since any linear order contains a cofinal well order.
Such spaces were called lobspaces and studied by S.W. Davis in Spaces with linearly ordered local bases, Topology proceedings 3, (1978), pp.3751.
6
Consider a totally ordered set $P$ on two elements $x<y$. Clearly it has the FPP. The interval topology on $P$ is discrete, so the map swapping $x$ with $y$ is continuous, but has no fixed point.
9
The answer is no.
A space is called resolvable if it contains two disjoint dense subspaces. Clearly $X$ is resolvable if and only if $\chi(X)=2$. Lets prove by induction on $n \geq 2$ that if $\chi(X) \leq n$ then $X$ is resolvable (and hence $\chi(X)=2$).
The base case $n=2$ is clear so suppose there is a coloring $f:X \to n+1$. Let $V$ be the union of ...
1
In Engelking's General Topology book, prop 2.6.11 gives an easy proof : if $\theta$ is surjective, then for any $g \colon A \to B^X \in (B^X)^A$ there is $h = \theta^{1}(g) \colon A \times X \to B \in B^{A \times X}$.
Now take $A \equiv B^X$ and $g \equiv Id_{B^X} \colon B^X \to B^X$, then $\theta^{1}(Id_{B^X}) = B^X \times X \to B = Ev$ is continuous.
4
I believe there is a regular nonsequential almost sequential space in ZFC+CH. (See below for a construction using a weaker assumption.)
For $S,T\subseteq \omega$ let $S\subseteq^* T$ denote inclusion modulo finite sets i.e. $S\setminus T$ is finite.
For $f,g:\omega\to\omega$ let $f\leq^* g$ denote dominance modulo finite sets i.e. $f(n)\leq g(n)$ except ...
0
See Corollary 2.8 in this paper:
If $X$ is perfect, compact and metrizable, then there is a nonatomic regular Borel measure of full support on $X$.
3
A simple solution: if $X$ is second countable, let $D=\{d_n : n =1,2,3,\ldots\}$ be a dense subset of $X$ and define $$\mu(A)= \sum_{n:d_n \in A}\frac{1}{2^n}$$ for all subsets of $X$.
Then clearly $\mu(X)=1$ and $\mu(O)>0$ for all $O$ nonempty and open.
If you want an atomless measure, we need at least that $X$ is crowded, and then we must maybe ...
2
During the night sleep my brain has found affirmative answers to both problems. The answer to the Problem is rather long, so I will present only the answer to Question, which is a bit tricky. First a definition.
A subset $D$ of a topological space $X$ is called $k$dense in $X$ if each compact subset $K\subset X$ can be enlarged to a compact set $\tilde K$...
2
Discussing this problem with Alex Ravsky we constructed the following
Example. The Euclidean topology $\tau_0$ on the set $\mathbb Q$ of rational numbers can be enlarged to a regular topology $\tau$ of weight $\omega_1$ such that the countable (and hence cosmic) topological space $(\mathbb Q,\tau)$ is not cometrizable.
The topology $\tau$ is constructed ...
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