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In Terrence Tao's article 245B notes 4: The Stone and Loomis-Sikorski representation theorems he gives a proof that not each sigma-complete Boolean algebra can be realized as a $\sigma$-complete Boolean algebra of sets. I have what appears to be a proof to the contrary, and I (and my colleagues) cannot find the error.

  • Every $\sigma$-complete Boolean algebra is a Boolean algebra
  • Every Boolean algebra admits a representation as a Boolean algebra of sets (Stone)
  • An isomorphism of Boolean algebras is an order isomorphism
  • An order isomorphism preserves all meets and joins present in its domain
  • Hence, a Boolean algebra isomorphism preserves all meets and joins in its domain
  • Therefore every if $f:B\to C$ is a Boolean algebra isomorphism and $B$ is $\sigma$-complete, $C$ is $\sigma$-complete and $f$ preserves countable joins and meets
  • Every $\sigma$-complete Boolean algebra admits a representation as a Boolean algebra of sets

What went wrong?

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The issue is that countable joins and meets in a Boolean algebra of sets need not be unions and intersections. So, the last step is wrong: an isomorphism from your Boolean algebra $B$ to an algebra of sets $C$ need not be a representation of $B$ as a Boolean $\sigma$-algebra, since it maps countable joins and meets in $B$ to countable joins and meets in $C$, which are not necessarily unions and intersections of sets.

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  • $\begingroup$ Oh! That is a subtle point! Thank you! $\endgroup$ – David Farrell Mar 14 at 5:20

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