# Tag Info

15

Sometimes a convenient substitute for Lebesgue measurability is the property of Baire. Just like Lebesgue measurability, the class of sets with this property is a $\sigma$-algebra containing the open subsets - indeed, open sets clearly have property of Baire, this class is closed under countable unions since meager sets are closed under countable unions, and ...

13

Alexandrov's gluing theorem: If one glues polygons together along their boundaries to form a closed surface homeomorphic to a sphere, such that no point has more than $2\pi$ incident surface angle, then the result is isometric to a convex polyhedron, uniquely determined up to rigid motions. There is as yet no effective procedure to actually construct the ...

6

Usha Kiran showed in An uncountable number of polar topologies and non-convex topologies for a dual pair that if the Mackey topology is distinct from the weak topology, then there are at least continuum-many distinct compatible topologies for the dual pair. The construction is based on unbounded subsets of the naturals, with equivalence given by having ...

6

The figure below gives a simple but extreme counterexample, which I think has all the lifting properties one might want except for actually being a true fibration. The map is the identity everywhere except for the linear segment with negative slope, which is mapped to the horizontal segment in the codomain. This map is a continuous bijection so all of the ...

5

Let us show how to find such a retraction for $n=2$ (I do not know if this method generalizes to higher dimensions). Given a compact set $C\subset\mathbb R^2$ and an open neighborhood $U\subseteq\mathbb R^2$ of $C$, choose a triangulation on $\mathbb R^2$ so fine that no triangle of the triangulation meets $C$ and $\mathbb R^2\setminus U$ simultaneously. ...

5

We can get a rather simple such function $f$. If if we measure this in terms of descriptive set theory, we get a Baire class 1 function (we need DST for Quasi-Polish spaces here, as $\mathcal{E} = \mathcal{O}(\mathbb{R})$ is not metric). If we want to do this in a weak axiom system, then $\mathrm{ACA}_0$ would do. Let us start by fixing some enumeration $(... 5 There are a number of easily stated problems in elementary computational geometry that have been solved only relatively recently, e.g., the origami existence theorem that a single rectangular sheet of paper can be folded into the shape of any connected polygonal region, even if it has holes; the fold-and-cut theorem that any shape with straight sides can be ... 5 Let$X = Y = S^1 = \mathbb{R}/ \mathbb{Z}$be the circle and let$f \colon X \to Y$be given by$f(t) = 2t$which is continuous and surjective. There exists no$g \in C(S^1,X)$such that$f \circ g = \mbox{Id}_{S^1}$, as$f$induces the doubling map on$\pi_1$. 5 In many cases,$f_\ast: C(Z,X)\to C(Z,Y)$,$g\mapsto f\circ g$is not surjective: Put$Z=Y$and$h=id_Y \in C(Z,Y)$, then$h$is in the range of$f_\ast$only if$f$has a right inverse. 4 No, this is not true. The simplest example is$f(z)=\int_0^ze^{-\zeta^2}d\zeta$. Preimage of the real line consists of infinitely many curves, each of them is mapped homeomorphically onto one or two intervals of the three intervals$(-\infty,-\pi/2),\; (-\pi/2,\pi/2),\; (\pi/2,+\infty)$. But none of the curves is mapped on the whole real line. The picture of ... 4 Note that$N$is homeomorphic to the union of$M$with$\DeclareMathOperator{\Cyl}{Cyl}$the mapping cylinder$\Cyl(\pi)$of the bundle projection$\newcommand{\pa}{\partial}\pi:\pa M\to X$. Denote by$M^\circ$the interior of$M$. Observe that the Poincare Duality for$M^\circ$(or equivalently for$(M,\pa M)$) implies$$H^{n-k}_{dR}(M)\cong H^k_{... 4 Choose a sequence$\varepsilon_n\to 0$and a$\varepsilon_n$-net$N_n$for each$n$. Assume$N_0$is a one-point set. For each point in$x\in N_k$choose a closest point in$y\in N_{k-1}$and connect$x$to$y$by a curve. Note that we can assume that diameter of the curve is at most$\delta_k$for a fixed sequence$\delta_k\to 0$. Consider the union$K'$... 4 With the axiom of choice: if$\kappa$is an infinite cardinal, then any collection of (at most)$\kappa$sets, each of cardinality (at least)$\kappa$, has an injective choice function. In this case$\kappa=2^{\aleph_0}$, and not only the collection of all nonempty open subsets of$\mathbb R$but even the collection of all uncountable Borel subsets of$\...

3

Doesn't this already fail in finite dimensions? Take $X = \mathbb{R}^2$ and let $A = B =$ the closed region bounded by $y = \sqrt{x^2 + 1}$ and $y = x + 1$, both for $x \geq 0$. The intersection with any ball is closed and therefore compact, but $A - B$ is not closed: it contains points arbitrarily close to $(0, -1)$ but does not contain that point itself.

3

The hyperstonean case can be dealt with using a result from Fremlin's Measure Theory. For every hyperstonean space $X$, we can find a semi-finite measure $\mu$ defined on the sets with the Baire property whose nullsets are exactly the meagre sets and which is inner regular with respect to compact subsets. Therefore $(X, \mathcal{BP}(X), \mu)$ (where $\... 2 EDIT: Pre-print available here. https://arxiv.org/abs/2005.01563 Note that I suspect that the use of the hypothesis of contractibility and the use of a homotopy is not really necessary, probably the hypothesis on the metric is enough. Consider the "generalised Cantor space", the carrier set being$2^{\omega_{1}}$and the basic open sets being sets ... 2 No, it need not. Take for example$X=Y=\mathbb{R}$, let$A=[0,1]$, let$B$be some nowhere dense perfect subset of$[2,3]$, and consider a Borel isomorphism$h:\mathbb{R}\cong \mathbb{R}$swapping$A$and$B$and fixing everything else. Then for each$U\subseteq A$the preimage$h^{-1}(U)$is meager in$\mathbb{R}$, hence has the property of Baire. Taking ... 2 Functions$\mathbb R^d\to \{0,1\}$are indicator functions$f=I_{A}$with$A=f^{-1}(\{1\})$and continuity with respect to the Sierpiński topology precisely means that$A$is open in$\mathbb R^d$. The topology of pointwise convergence on$C(\mathbb R^d,\{0,1\})$is strictly coarser than the compact-open topology: Since the only interesting open set in$\{0,...

2

Paracompact non-Hausdorff spaces can admit partitions of unity, but you need a stronger property than paracompactness. A cover admits a subordinate partition of unity iff it has a locally finite refinement by cozero sets. A cozero set of a space $X$ is a set of a the form $f^{-1}(0,1]$, for a continuous function $f: X \to [0,1]$. For a space, every open ...

2

This is not an answer, but I put it here to stop close votes that are apparently based on a misreading of the question. I do not know the answer to the OP's question, but I will point out that both hypotheses are necessary. For example, drop the hypothesis that $A$ is convex. Then there is a counterexample with $A=B$; namely, $A=\bigcup_n \{(n+1)e_n, ne_n\}$...

2

Yes, this is known and you don't need local compactness even as long as $X$ is completely regular (locally compact spaces are completely regular). I think that the best way to prove is to go via the Čech–Stone functor, which is available precisely for completely regular spaces. Consider $\beta X$ and note that $K, W\subset X\subset \beta X$ remain compact ...

2

In the case that $G$ is discrete, what you've defined last as $\bar{d}(A)$ is usually considered the upper (Banach) density of $A$. Moreover, in this case $\bar{d}(A)$ is the supremum of $\bar{d}_{\mathcal{F}}(A)$ over all Folner sequences $\mathcal{F}$ (or nets in the uncountable case). Recall also that a discrete group $G$ is amenable if and only if it ...

2

1) The set of continuous everywhere but differentiable nowhere functions on the unit interval is a meagre set of measure 1. 2) The existence of a space filling curve, or more generally surjective continuous maps $S^m \to S^n$ for $n>m$ (and then the fact that however any such map is homotopic to a map that misses a point).

2

The answer is negative. In $X :=\ell_2 \oplus_\infty \ell_1$ let $x_n = (M_n \delta_n, 2 e_n)$ and $y_n = (M_n \delta_n, e_n)$, where $(\delta_n)$ is the unit vector basis for $\ell_2$ and $(e_n)$ is the unit vector basis for $\ell_1$ and $M_n \uparrow \infty$ fast. Let $B$ be the convex hull of the $x_n$ and $y_n$ and let $A$ be the closure of $B$. Then $... 2 For me the most basic compact spaces are$\{x_n: n \in \Bbb N\} \cup \{x\}$when$x_n \to x$, (the countable cofinite space is a special case), of course all finite spaces, and all ordered topological spaces with a suprema for all subsets ("very order complete", usual order completeness being: "every nonempty subset that is bounded above has a supremum"). (... 2 Elaborating on Michael Greinecker's comment: if$X$is a compact Hausdorff space then the map$i \colon X \to \Pi_{C(X,I)} I$given by$i(x)_f = f(x)$, where$I = [0,1]$, is a homeomorphism onto its image. So every compact Hausdorff space can be expressed as the closed subspace of the product of closed unit intervals, which is a partial negative answer to ... 1 These are many questions... Let me call this E-structure. The first part of Question 1 has a positive answer. Indeed, the 7-sphere admits a magma structure$(x,y)\mapsto xy$for which left and right multiplications are invertible with continuous inverse (octonion multiplication). Hence it admits and E-structure with$\lambda(x,y)=xy^{-1}$and$\rho(x,y)=x^{...

1

These are a few of the existence-type results that may be of interest to you. $\text{Green, Tao (2006)}$: There are arbitrarily long arithmetic progressions in primes. The proof they provided is based on extending Szemerédi's Theorem [1975, Endre Szemerédi] which, in itself, is an existence-type result, stating that any subset of $\mathbb{Z}$ with positive ...

1

adding other great examples, many of them provided in the comments section (16) Kakeya needle problem and Besicovitch sets: You want to rotate a needle of unit length by $360°$. What is the region with smallest area to do that? It turns out there is no lower bound > 0 for the area of such a region, i.e. you can find arbitrarily small such regions. (https://...

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This is meant to fill in some of the details outlined by Anton Petrunin's answer, and also to refine the statement slightly. Recall that a compact connected Hausdorff space is called a continuum. We will call a topological space $X$ continuum-connected if every $x,y\in X$ can be joined by a continuum, i.e. there is a continuum $K\subset X$ that contains ...

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