Timeline for Are two forms of the Dual Schroeder-Bernstein property equivalent?

11 events

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 15, 2015 at 5:53	vote	accept	Yaakov Baruch
Jan 15, 2015 at 0:57	history	edited	Asaf Karagila♦		edited tags
Jan 15, 2015 at 0:52	answer	added	Asaf Karagila♦		timeline score: 7
Jan 14, 2015 at 21:56	history	edited	Asaf Karagila♦	CC BY-SA 3.0	Assaf is someone else.
Jan 14, 2015 at 21:54	history	edited	Yaakov Baruch	CC BY-SA 3.0	added 73 characters in body
Jan 14, 2015 at 21:51	comment	added	Asaf Karagila♦		This is a difficult question, since we don't know any explicit models where $\sf ZF+\lnot AC+DSB$ hold, or $\sf ISB$ for that matter. At best we have "local examples" (where the general principle fails, but nontrivially holds for some set). My guess is that the implication is false.
Jan 14, 2015 at 21:46	comment	added	Yaakov Baruch		Does ZF+ISB $\implies$ DSB? Should I edit?
Jan 14, 2015 at 21:45	comment	added	Asaf Karagila♦		As for your last remark, this thread shows that $\sf ISB$ is not provable in $\sf ZF$.
Jan 14, 2015 at 21:44	comment	added	Asaf Karagila♦		What do you mean "the converse"?
Jan 14, 2015 at 21:37	history	asked	Yaakov Baruch	CC BY-SA 3.0