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Sep 27, 2016 at 11:54 comment added user97621 Thank you! Yes, this appears an interesting counterexample, though not exactly to what I was asking (there was no requirement that $A=A_1\cup A_2$ where $A_1$ and $A_2$ are disjoint).
Sep 27, 2016 at 11:47 vote accept user97621
Sep 16, 2016 at 14:59 history edited Taras Banakh CC BY-SA 3.0
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Sep 16, 2016 at 14:52 history undeleted Taras Banakh
Sep 16, 2016 at 14:07 history deleted Taras Banakh via Vote
Sep 16, 2016 at 14:04 history answered Taras Banakh CC BY-SA 3.0