Timeline for Is it consistent with ZFC that the real line is approachable by sets with no accumulation points?

6 events

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Jun 28 at 1:24	history	edited	Pedro Lauridsen Ribeiro	CC BY-SA 4.0	verb repetition removed
Jun 27 at 21:02	comment	added	Joel David Hamkins		If you press my argument harder, you get the theorem: every lineared ordered family of sets of reals, each set in the family having no accumulation point, has a countable union. If it had an uncountable union, then that union would have an accumulation point, and so some countable set of points in the union would have an accumulation point, but they would have to all appear by some stage, or else the cofinality would be countable and the union would be countable.
Jun 27 at 16:17	comment	added	Joel David Hamkins		Could argue directly, without mentioning $\omega_1$, like this: every rational appears in some $s\in S$, but they cannot appear together, so $S$ has a cofinal $\omega$ sequence. But since each $s$ is countable, the union of all of them will be countable, so it isn't $\mathbb{R}$.
Jun 27 at 15:58	history	edited	Joel David Hamkins	CC BY-SA 4.0	added 237 characters in body
Jun 27 at 15:28	vote	accept	Julian Newman
Jun 27 at 15:25	history	answered	Joel David Hamkins	CC BY-SA 4.0