Timeline for On a property possibly separating countable and not countable cardinals

7 events

when toggle format	what		by	license	comment
Jun 9, 2020 at 8:33	vote	accept	Dieter Kadelka
Jun 8, 2020 at 23:31	answer	added	bof		timeline score: 8
Jun 8, 2020 at 18:32	comment	added	Dieter Kadelka		Thank you bof and @Ashutosh. I'm sorry, but I prefer the answer of bof. It is simple and effective. Can you make it to an answer?
Jun 8, 2020 at 16:10	comment	added	bof		@Ashutosh Todorcevic's result is much stronger, but you don't need Todorcevic to get $f$ with the property you need, it's just an exercise in elementary set theory. For each $\alpha\in\omega_1$ choose an injection $\psi_\alpha:\alpha\to\mathbb N$. For $\beta\lt\alpha\in\omega_1$ define $f(\alpha,\beta)=f(\beta,\alpha)=\psi_\alpha(\beta)$. So $f:\omega_1\times\omega_1\to\mathbb N$, and $f$ is unbounded on $X\times X$ if $X\subseteq\omega_1$ is uncountable, indeed, if $X$ has order type $\ge\omega+1$.
Jun 8, 2020 at 7:30	comment	added	Ashutosh		Suppose $c:[\omega_1]^2 \to \omega$ satisfies: For every uncountable $Y \subseteq \omega_1$, the range of $c \upharpoonright [Y]^2$ is unbounded in $\omega$. The existence of such $c$ and much more is well known (See S. Todorcevic, Partitioning pairs of countable ordinals, Acta Math., 159(3–4):261–294, 1987). So you can define $f(x, y) = c(\{x, y\})$ if $x \neq y$, and $0$ otherwise. It follows that there is such an $f:X^2 \to \omega$ iff $X$ is uncountable.
Jun 8, 2020 at 0:48	history	edited	YCor		edited tags
Jun 8, 2020 at 0:40	history	asked	Dieter Kadelka	CC BY-SA 4.0