Timeline for Order-embeddability of ${\frak b}$ and ${\frak d}$ in $\mathbb{R}$ [duplicate]

Current License: CC BY-SA 4.0

11 events

when toggle format	what		by	license	comment
May 16, 2023 at 18:19	vote	accept	Dominic van der Zypen
May 16, 2023 at 17:10	comment	added	Noah Schweber		@YCor Note that as written that approach actually requires choice (it's consistent with ZF that there is an unbounded subset of the reals all of whose countable subsets are finite). It's better to just pick the lex-least rational between successive elements in the range of a putative embedding.
May 16, 2023 at 16:54	comment	added	Noah Schweber		The answer to your earlier question applies here as well.
May 16, 2023 at 16:54	history	closed	YCor Noah Schweber lo.logic		Duplicate of Smallest ordinal $\mu$ not embeddable in ${\cal P}(\omega)$
May 16, 2023 at 16:40	comment	added	YCor		If an infinite limit ordinal embeds into $\mathbf{R}$, its convex hull is order-isomorphic to $\mathbf{R}_{\ge 0}$, so changing the embedding we can suppose that its convex hull is $\mathbf{R}_{\ge 0}$. Now assuming the ordinal is $\omega_1$, we obtain an unbounded subset of $\mathbf{R}_{\ge 0}$ in which every countable subset is bounded. This is of course absurd.
May 16, 2023 at 16:30	answer	added	Arno		timeline score: 1
May 16, 2023 at 16:22	comment	added	Alessandro Codenotti		The same is true for $\Bbb R$
May 16, 2023 at 16:20	history	edited	Dominic van der Zypen	CC BY-SA 4.0	edited body; edited title
May 16, 2023 at 16:19	comment	added	Dominic van der Zypen		Right - will modify, I should have written \mathbb{R}
May 16, 2023 at 16:18	comment	added	Alessandro Codenotti		Isn't every ordinal embeddable in $\Bbb Q$ countable? What am I missing?
May 16, 2023 at 16:15	history	asked	Dominic van der Zypen	CC BY-SA 4.0