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Nov 9, 2019 at 9:03 comment added YCor BTW if $G$ is non-amenable and countable, then there is a countable $G$-invariant subalgebra of $2^G$ with no invariant mean. Also there is a countable non-atomic $G$-invariant subalgebra of $2^G/$(fin) with no invariant mean.
Nov 9, 2019 at 8:45 vote accept Colin Reid
Nov 9, 2019 at 1:29 history answered YCor CC BY-SA 4.0