# Representation of sigma complete Boolean algebras

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?

• – Eric Wofsey Mar 14 at 4:48

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.