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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 ...


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 ...


8

Note that replacing "well-ordered" by "linearly-ordered" produces an equivalent property since any linear order contains a cofinal well order. Such spaces were called lob-spaces and studied by S.W. Davis in Spaces with linearly ordered local bases, Topology proceedings 3, (1978), pp.37-51.


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, 1645-1653 (2012). ZBL1241....


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 $k\mapsto k-n+m$ which takes $n$ to $m$...


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.


5

The number of compact Hausdorff group topologies on a given group strongly depends on the algebraic structure of the group. For example, any finite-dimensional 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 ...


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 ...


4

$2^\kappa$ for uncountable $\kappa$ is a ZFC example of an almost sequential non-sequential 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)| < \...


4

I believe there is a regular non-sequential 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 ...


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 non-distortion 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 ...


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 Scott-closed ideals. If an ideal ...


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....


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.


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 "$...


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 $O$ non-empty 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 ...


2

The answer is NO, i.e. the following non-singleton 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{1-x^2})\,:\ -1<x<1\} $ - PROOF: Let $\ X'\subseteq X\ $ be homeomorphic to $\ X.\ $ The every ...


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 ...


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 "absorbing-bounded" (Because it contains a product basic open set which has almost ...


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 ...


1

There is nothing to say in the general case, but in the case of a continuous dcpo $D$ there is a well-known 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 $...


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 $x\neq 0$?


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.


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 co-countable topology is a topological space $(X,\mathcal{T})$ where $X$ is an ...


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 I-III, we expect the readerto be familiar with the concept of a smooth function and a complex ...


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