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Aug 1, 2013 at 18:15 comment added Igor Khavkine You're right. I was a bit sloppy. I've edited to correct that.
Aug 1, 2013 at 18:14 history edited Igor Khavkine CC BY-SA 3.0
Fixed typo.
Aug 1, 2013 at 14:55 comment added user31967 So by $D[F]$ in last line you don't mean $D[F]$ you mean $\{D[a] \mid a\in F\}$!
Aug 1, 2013 at 14:36 comment added Igor Khavkine In the usual way, $\sigma_D = \{ D(x) \mid x\in X \}$. And sorry, I don't have a counterexample. I only looked at the general argument.
Jul 31, 2013 at 23:16 vote accept CommunityBot
Jul 31, 2013 at 23:06 comment added user31967 How is $\sigma_D$ defined? And do you have the counterexample in last part?
Jul 31, 2013 at 20:11 history answered Igor Khavkine CC BY-SA 3.0