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I'm not sure if this is research level, so feel free to vote to migrate.

Suppose we have a complete boolean algebra $A$, with a dense, $\sigma$-complete subalgebra $B$, and a $\sigma$-complete homomorphism $h : B \to C$, where $C$ is complete. Does $h$ have a $\sigma$-complete extension $h' : A \to C$?

EDIT: I forgot to say that I'm interested in the case that $A$ is atomless.

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2 Answers 2

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If the Boolean algebra $A$ is c.c.c., then of course every $\sigma$-complete dense subalgebra $B$ is all of $A$, and so in this case the answer is trivially affirmative.

If $A$ is not c.c.c., however, then fix an uncountable maximal antichain $X\subset A$, which we may assume has size $\aleph_1$, and proceed along Joseph's idea. Namely, let $B\subset A$ consist of the elements $a\in A$ that are below the join of a countable subset of $X$, or above the complement of such a join. This is a $\sigma$-complete subalgebra of $A$, which is dense because it has every element below any element of $X$. Let $h:B\to C=\{0,1\}$ map $a$ to $0$ when it is below the join of a countable subset of $X$, and otherwise $h(a)=1$. This is a $\sigma$-complete homomorphism, which corresponds to the co-countable filter on the subsets of $X$, as in Joseph's answer.

If we can extend this $h$ to a $\sigma$-complete homomorphism $h':A\to\{0,1\}$, however, then the pre-image of $1$ will give a $\sigma$-complete nonprincipal ultrafilter on $X$, which is impossible, since $\aleph_1$ is not a measurable cardinal.

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  • $\begingroup$ For B to be dense in A, you probably also need X to be a maximal antichain in A. An example of such an A would be the boolean completion of $\mathcal{P}(\omega_2) \text{ mod }[\omega_2]^{\leq \omega}$. $\endgroup$
    – Ashutosh
    Commented Mar 28, 2014 at 20:53
  • $\begingroup$ Oh yes, that is what I meant, but I see I forgot to say "maximal". I have edited. $\endgroup$
    – Joel David Hamkins
    Commented Mar 28, 2014 at 20:55
  • $\begingroup$ And there are numerous examples of non-c.c.c. complete Boolean algebras arising in forcing, such as any forcing notion collapsing cardinals nontrivially. $\endgroup$
    – Joel David Hamkins
    Commented Mar 28, 2014 at 21:01
  • $\begingroup$ Yes :). I was about to type the power set of $\omega_2$ mod countable example when I saw your much more general argument. $\endgroup$
    – Ashutosh
    Commented Mar 28, 2014 at 21:07
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    $\begingroup$ For each non-c.c.c algebra $A$, we can set $C$ to be any complete c.c.c. Boolean algebra and the same result with the same argument will work except that instead of having a $\sigma$-complete ultrafilter on $\aleph_{1}$, we would have a $\sigma$-complete $\sigma$-saturated ideal on $\aleph_{1}$ which is still impossible. In other words, the image $C$ does not have to be trivial. $\endgroup$
    – Joseph Van Name
    Commented Mar 28, 2014 at 21:55
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No. Let $X$ be an uncountable set whose cardinality is below the first uncountable measurable cardinal. Let $B$ be the collection of all countable and cocountable subsets of $X$. Then $B$ is a dense $\sigma$-subalgebra of the power set algebra $P(X)$. Let $h:B\rightarrow\{0,1\}$ be the homomorphism where $h(R)=0$ for countable $R$ and $h(R)=1$ for cocountable $R$. Then $h$ has no extension to a $\sigma$-complete homomorphism $h':P(X)\rightarrow\{0,1\}$, otherwise $(h')^{-1}[\{1\}]$ would be non-principal a $\sigma$-complete ultrafilter on $X$, a contradiction.

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  • $\begingroup$ Thanks. What about atomless $A$? $\endgroup$
    – Monroe Eskew
    Commented Mar 28, 2014 at 18:00

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