# 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 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 2021/11/28 01:31:30 CET