Timeline for If a topological space X has $\aleph_1$-calibre, then it must be star countable?

11 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Nov 17, 2011 at 5:40	vote	accept	Paul
Nov 16, 2011 at 2:48	vote	accept	Paul
Nov 16, 2011 at 2:49
Nov 16, 2011 at 0:29	comment	added	Joel David Hamkins		Henno, I edited to give the proof.
Nov 16, 2011 at 0:28	history	edited	Joel David Hamkins	CC BY-SA 3.0	Added theorem on products of calibre aleph_1 spaces; added 2 characters in body
Nov 15, 2011 at 18:14	comment	added	Henno Brandsma		@Joel, indeed the productivity of calibre $\aleph_1$ spaces is well-known.
Nov 15, 2011 at 1:51	comment	added	Joel David Hamkins		In the paper to which I linked, we have $\kappa\geq 2^{c^+}$, in which case $\mathbb{R}^\kappa$ is similarly large. But if the $\mathbb{N}^{\omega_1}$ example works out, it could still be $\leq 2^{\aleph_0}$, since $2^{\aleph_0}=2^{\aleph_1}$ is consistent with ZFC. So I don't know.
Nov 15, 2011 at 1:49	comment	added	Paul		What if the cardinality of the topological space X is additionally < = $2^{\aleph_0}$?
Nov 15, 2011 at 1:31	comment	added	Joel David Hamkins		It seems that the $\Delta$-system argument shows that the $\aleph_1$-calibre spaces are closed under arbitrary products.
Nov 15, 2011 at 0:58	history	edited	Joel David Hamkins	CC BY-SA 3.0	deleted 38 characters in body
Nov 15, 2011 at 0:43	history	answered	Joel David Hamkins	CC BY-SA 3.0