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Jun 28, 2019 at 7:02 vote accept LJGC
Jun 27, 2019 at 16:14 comment added Robert Furber @YCor If you replace Borel by Baire, your statement is correct (and can be found in Halmos's book, for example). However, it is false with Borel if $E$ is uncountable - a singleton in $2^E$ is a Borel set that is not a Baire set.
Feb 7, 2019 at 22:03 comment added YCor Given the free $\sigma$-algebra $A_E$ on $E$ with free generators $(x_t)_{t\in E}$, is it correct that mapping $A_E$ to $2^{2^E}$ by $x_t\mapsto \{Y\in 2^E:t\in Y\}$ induces an isomorphism from $A_E$ onto the $\sigma$-algebra of Borel subsets of the compact space $2^E$? Surjectivity is clear, I'm just not sure for injectivity. If it indeed holds, this is a pretty concrete description (from which the cellularity fact is obvious).
Feb 5, 2019 at 20:25 answer added Ramiro de la Vega timeline score: 5
Feb 4, 2019 at 18:18 history asked LJGC CC BY-SA 4.0