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Mar
21
comment For a partition of $\mathbb{R}$ into countably infinite sets, must there be an almost-disjoint family of $2^{\frak c}$ many selectors?
Under GCH this is Problem 19/A in the old "Unsolved problems in set theory" paper of P. Erdős and A. Hajnal.
Mar
13
comment Communal problem books
I believe there is (or was) a problem book at the University of Calgary, when Eric Milner was there.
Mar
13
comment Characterising subsets of the reals as ordered spaces
@JoelDavidHamkins Oh, right. Thanks for correcting me.
Mar
13
comment Characterising subsets of the reals as ordered spaces
I think 2 can be put more simply: there is a countable $Q\subseteq L$ such that $a,b\in L,a\lt b\implies[a,b)\cap Q\ne\emptyset.$ Let $Q=\{q_n:n\lt\omega\}$ and define $f:L\to\mathbb R$ by $f(x)=\sum_{q_n\lt x}2^{-n}$. Or something like that.
Mar
13
comment Characterising subsets of the reals as ordered spaces
Doesn't "linear order with a countable dense set" do the trick?
Mar
11
comment Generalization of Hamiltonian cycle
Anyway, since the OP counts $K_2$ as a cycle, there is no need for infinite cycles in the partition, unless one has qualms about the axiom of choice.
Mar
11
comment Generalization of Hamiltonian cycle
Since the OP defined "neighborly" with an injection rather than a bijection, the components of an infinite "neighborly" graph may have vertices of degree one.
Mar
5
comment “For sufficiently large” vs. “For all sufficiently large”
@GerhardPaseman The existential interpretation seems implausible to me. What is the "sufficiently large" doing in "for some sufficiently large $x$"? What's the difference between "$P(x)$ holds for some $x$" and "$P(x)$ holds for some sufficiently large $x$"? I guess the latter must mean $\exists N\ \exists x\ge N\ P(x)$? Seems kind of silly, doesn't it?
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
I believe I've seen quasidisjoint family used as a synonym for $\Delta$-system, i.e., a family of sets in which each pair has the same intersection.
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
@AsafKaragila: Choose $a,b\in X,\ a\ne b.$ Define $f:X\times X\to S$ as follows. Let $f(x,x)=\{x\}.$ Let $f(a,b)=\emptyset.$ If $x\ne y$ and $(x,y)\ne(a,b),$ let $f(x,y)$ be the unique member of $S$ containing $\{x,y\}$ if there is one, otherwise let $f(x,y)=\emptyset.$ The range of $f$ is a superset of $S$.
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
@AsafKaragila: Let $A=\mathbb R$ and $B=\mathbb R\cup\omega_1$. There is a surjection from $A$ to $B$ but it's consistent that $|A|\lt|B|$.
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
@AsafKaragila: This choiceless stuff always confuses me. Does $|A|\lt|B|$ imply that there is no surjection from $A$ to $B$?
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
@AsafKaragila: Well then, that answers Dominic van der Zypen's question, doesn't it? If $S$ is a "quasi-disjoint" (as defined here) family of subsets of $X$, then there is a surjection from $X\times X$ to $S$, right?
Mar
4
comment Quasi-disjoint subsets of an infinite set and $\neg \mathsf{AC}$
As a weaker version of your question, is it consistent that there exist an infinite set $X$ and a surjection from $X\times X$ to $\mathcal P(X)$?
Mar
4
comment a question about Brooks' Theorem for $\Delta =4$
It's Brooks's theorem, not Brook's.
Feb
23
awarded  Necromancer
Feb
22
comment Translates of null sets
@MohammadGolshani: Was that comment addressed to me? If so, what is your point? It suffices that every $G_\delta$ null set is covered by the union of countably many translates of $N$, and there are only continuum many $G_\delta$ null sets. The fact that there are many more null sets seems irrelevant.
Feb
22
comment Translates of null sets
@MarioCarneiro: I am mystified. What's the difference between "every null set is covered by countably many translates of N" (Null's question) and "every null set is a subset of a countable union of translates of N" (the question I asked)?
Feb
22
comment Translates of null sets
@MarioCarneiro Looks like the same question to me. What is the difference?
Feb
21
comment Translates of null sets
This answer (to a somewhat different question) on Math Stack Exchange sketches what is claimed to be a proof that no such $N$ exists.