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Apr 9, 2023 at 16:50 history edited LSpice CC BY-SA 4.0
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Apr 9, 2023 at 16:35 history edited YCor CC BY-SA 4.0
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Mar 24, 2012 at 20:33 comment added Misha @Guntram: You are right of course, but if you eliminate the (uncountable) Axiom of Choice, then, at least in Solovay's model, all subsets of ${\mathbb R}$ are measurable and we are in the situation discussed in mathoverflow.net/questions/64116/… (see pm's answer).
Mar 24, 2012 at 10:00 comment added Guntram For the existence of wild automorphisms of $\mathbf{C}$ and for the construction Misha quoted (due to Borel--Tits) you need the axiom of choice, however.
Mar 23, 2012 at 14:42 vote accept Hugo Chapdelaine
Mar 23, 2012 at 14:39 comment added Hugo Chapdelaine Right, good point!
Mar 23, 2012 at 12:37 comment added YCor You can embed $SL_2(\mathbf{R})$ into $SL_2(\mathbf{C})$ and twist by a wild automorphism of $\mathbf{C}$.
Mar 23, 2012 at 12:34 answer added paul garrett timeline score: 2
Mar 23, 2012 at 11:45 answer added Misha timeline score: 9
Mar 23, 2012 at 11:01 history edited Marc Palm
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Mar 23, 2012 at 10:52 answer added Marc Palm timeline score: 0
Mar 23, 2012 at 10:26 history asked Hugo Chapdelaine CC BY-SA 3.0