Factoids #3

2021-07-10, post № 246

mathematics, #topology, #reals, #lipschitzian

X) An intriguingly delicate arbitrarily small countable open cover of uncountably many unbounded points inferred from locally lipschitzian maps’ inability to increase a set’s Hausdorff dimension

Claim. For any uncountable Lebesgue measure zero set Z\subset\mathbb{R}, there exists a point z_0\in Z such that \forall\,\varepsilon>0:{]}z_0,z_0+\varepsilon{[} is still uncountably infinite. Define Z_>:=Z\cap{]}z_0,\infty{[} and on this right portion u:Z_>\to\mathbb{R} as u(z):=\sfrac1{(z-z_0)}. Looking at the graph \mathrm{graph}\,u\subset\mathbb{R}^2, it is a Hausdorff measure zero set — meaning the entirety of all uncountably many points clustered at z_0 span an infinite vertical distance and yet are coverable by countably many sets of arbitrarily small diameter in sum.
Proof. Since \sfrac1x is locally lipschitzian and the reals are Lebesgue \sigma-compact, its image cannot increase the one-dimensional Hausdorff measure which is presumed to vanish.

XI) A closed set in the plane whose projection onto the first coordinate is open in the reals

Claim. There exists a set C\subset\mathbb{R}^2 which is closed whilst its projection under \pi:\mathbb{R}^2\twoheadrightarrow\mathbb{R},(x,\_)\mapsto x is open, more precisely \pi(C)={]}0,1{[}.
Proof. Define C:=\bigcup_{j\geq 2}\overline{\mathrm{B}_{\sfrac12-\sfrac1j}}((\sfrac12,j-2)) and note that this union of closed balls with pairwise positive distance is again closed via sequence closedness. Furthermore, \pi(C)\subset{]}0,1{[} is apparent and for any 0<x<1 there exists an index j\geq 2 such that x\in{[}\sfrac1j,1-\sfrac1j{]}={[}\sfrac12-(\sfrac12-\sfrac1j),\sfrac12+(\sfrac12-\sfrac1j){]} and thus x\in\pi(C) since it lies in the projection of a closed ball horizontally centered at a half with radius \sfrac12-\sfrac1j.

XII) Characterizing separability by subsequence approximabilty

Claim. A general topological space (X,\tau) is separable iff there exists a sequence q_j\in X such that for any x\in X there exists a subsequence (q_j')\subset(q_j) which satisfies q_j'\sim_{\mathrm{lim}}x.
Proof. “⇒”. Let Q\subset X be countable and dense. Let q:\mathbb{N}\to X be a (global) enumeration, its evaluation denoted by q_j. Since Q lies densely, \forall\,\emptyset\neq O\in\tau\,\exists\,j\in\mathbb{N}:q_j\in O holds. For arbitrary point x\in X and open neighborhood x\in O_x\in\tau follows the existence of a q_j\in O_x and thus q_j'\sim_{\mathrm{lim}}x for a subsequence.
“⇐”. Define Q:=q(\mathbb{N})\subset X. This set is by definition countable and for arbitrary x\in X there exists (q_j^x)\subset(q_j) which approximates q_j^x\sim_{\mathrm{lim}}x. Following, for an open neighborhood x\in O\in\tau an index j_O exists with q_{j_O}^x\in O, showing O\cap Q\neq\emptyset and thereby that Q lies dense.

XIII) Discontinous inclusion

Claim. On a set X, endow two nontrivially progressively fine topologies \tau\subsetneq\tau'\subset\wp(X). Embedding \iota:(X,\tau)\hookrightarrow(X,\tau') is a discontinuous undertaking.
Proof. For arbitrary open set O\in\tau'-\tau\neq\emptyset, it follows \iota^{-1}(O)=O\notin\tau, thus \iota is discontinuous.
Claim. Any non-discrete topology \tau\subsetneq\wp(X) endowed on a set X can be embedded into a finer topology on the same set such that this inclusion is discontinuous.
Proof. From \tau\neq\wp(X), one gets the existence of a point x\in X whose singleton \{x\}\notin\tau is not open. Refining \tau':=\sigma_X(\tau+\{\{x\}\}) where \sigma_X denotes the generated topology together with the previously shown yields the claim.
Factoid XIII) was originally formulated June 2020.
Jonathan Frech's blog; built 2021/09/06 04:50:48 CEST