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Jan 23, 2015 at 14:41 comment added MikeTeX Yes, actually, my proof is wrong. But maybe it is now not too difficult to find a counter example.
Jan 23, 2015 at 14:33 comment added Eric Wofsey It is not necessarily the case that $e(\downarrow \alpha)=\downarrow e(\alpha)$; only $\subseteq$ holds in general. If instead of $e(s_i)$ and $e(t_j)$ you mean things like $\downarrow e(\alpha)$, then they will still cover $Q$ and you get two disjoint open sets in $Q$, but you can't be sure that these sets contain $x'$ and $y'$.
Jan 23, 2015 at 14:14 history edited MikeTeX CC BY-SA 3.0
added 18 characters in body
Jan 23, 2015 at 14:09 history answered MikeTeX CC BY-SA 3.0