# Tag Info

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....

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 ...

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 "absorbing-bounded" (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$k\mapsto k-n+m$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 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$...

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 ...

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

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 Scott-closed ideals. If an ideal ...

4

 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 ...

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 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 $A=\{x \in 2^\kappa: |x^{-1}(1)| < \... 3 No. Just looking at countably many generators we can produce a continuum of pairwise disjoint clopen subsets of$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 I-III, we expect the readerto be familiar with the concept of a smooth function and a complex ... 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. 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 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 ... 0 See Corollary 2.8 in this paper: If$X$is perfect, compact and metrizable, then there is a non-atomic 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$O$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 ...

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