bio | website | |
---|---|---|
location | ||
age | ||
visits | member for | 1 year, 4 months |
seen | 7 hours ago | |
stats | profile views | 421 |
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. |