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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 ...
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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 ...
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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 ...
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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 ...
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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.
...
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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 $(...
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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 ...
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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 $\pi_1$.
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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.
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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 ...
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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_{...
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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 $K'$ ...
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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.
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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 $\...
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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 ...
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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 ...
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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,...
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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 ...
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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, $A=\bigcup_n \{(n+1)e_n, ne_n\}$...
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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 ...
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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 ...
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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).
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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 $...
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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"). (...
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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 ...
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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^{...
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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 ...
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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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