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I'm reading the Tarski's paper "Axiomatic and algebraic aspects of two theorems on sums of cardinals". At page 99, he states the following theorem.

Theorem A Boolean $\sigma$-algebra is isomorphic to a $\sigma$-algebra of sets if and only if all its elements $\neq1$ are contained in a prime $\sigma$-ideal (i.e., a maximal ideal containing the sup of any countable family of its elements).

The necessity seems obvious. As for the sufficiency, Tarski says that it can be proved in the same way one can prove the Stone representation theorem. Unfortunately, I do not understands how. Do you have any suggestion on the matter?

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The proof is nearly identical to the proof of the Stone representation theorem; you just add some $\sigma$s in appropriate places. Let $B$ be a Boolean $\sigma$-algebra and let $X$ be the set of prime $\sigma$-ideals in $B$. Define $f:B\to P(X)$ by $f(b)=\{I\in X:\neg b\in I\}$. Then $f$ is a homomorphism, and is in fact a $\sigma$-homomorphism since each $I\in X$ is a $\sigma$-ideal. Now if your hypothesis holds, then $f$ is injective, since $f(b)=\emptyset$ means that $\neg b$ is not in any prime $\sigma$-ideal so $\neg b=1$ and $b=0$. Thus, $B$ is isomorphic to the $\sigma$-algebra of sets given by the image of $f$.

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  • $\begingroup$ @Eric_Wofsey Can you explain why $f$ is a $\sigma$-homomorphism? $\endgroup$
    – W4cc0
    Commented Oct 24, 2020 at 16:35
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    $\begingroup$ What part of the definition are you having trouble verifying? It may help to observe that $\neg b\in I$ iff $b\not\in I$ since $I$ is prime. $\endgroup$ Commented Oct 24, 2020 at 19:18
  • $\begingroup$ Ok, looking at them in terms of functions it seems more clear! $\endgroup$
    – W4cc0
    Commented Oct 25, 2020 at 16:10

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