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All countable atomless algebras are isomorphic. Can one give an example of a pair of mutually non-isomorphic atomless Boolean algebras of cardinaliy continuum?

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I'm glad that you were happy with my answer but I suggest that you wait a little bit before accepting it (maybe a day or two) because others might have a lot more to say than I do and accepting an answer generally discourages others to look at questions and/or write up answers. – t.b. May 22 '12 at 19:21
Well, at least I can verify ccc condition in the algebras you mentioned, so this is fairly good example! – MarkNeuer May 22 '12 at 19:56
up vote 5 down vote accepted

Take $\mathfrak{A}$ to be the Lebesgue measure algebra and $\mathfrak{B} = P\mathbb{N}/[\mathbb{N}]^{\lt \omega}$, the quotient algebra of $P\mathbb{N}$ modulo the ideal of finite sets. Then both are atomless and have cardinality continuum but they are not isomorphic because $\mathfrak{A}$ is ccc while $\mathfrak{B}$ isn't.

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Did you mean the Borel measure algebra? The Lebesgue one is strictly larger than $\mathcal P(\mathbb N)/\mathrm{Fin}$. – Asaf Karagila Dec 29 '12 at 23:12
For me the measure algebra of a measure space is the algebra of measurable sets modulo the ideal $\mathcal{N}$ of null sets. Thus, $\mathfrak{A} = \mathfrak{L}/\mathcal{N} \cong \mathfrak{B} / (\mathfrak{B} \cap \mathcal{N})$ has cardinality $\mathfrak{c}$. Otherwise ccc wouldn't tell $\mathfrak{A}$ and $\mathfrak{B}$ apart, would it? :) – t.b. Feb 3 '13 at 9:19
Hmm, that's the second time someone mentions this to me recently. – Asaf Karagila Feb 3 '13 at 9:58
I suppose it's a matter of background and culture. I've seen reduced measure algebra used for the same thing (I think Givant/Halmos use that). However, it doesn't make much sense to me to use the term measure algebra as a synonym of $\sigma$-algebra and from this point of view $\Sigma/\mathcal{N}$ is pretty much the only possible interpretation. It seems that for analysts $\Sigma/\mathcal{N}$ is the interesting thing to consider, not $\Sigma$ itself. – t.b. Feb 3 '13 at 10:12

Let $\langle X,\tau\rangle$ be a topological space. A set $U\in\tau$ is a regular open set iff $\operatorname{int}_X\,\operatorname{cl}_XU=U$. Let $\mathrm{RO}(X)$ be the family of regular open subsets of $X$. For $U,V\in\mathrm{RO}(X)$ define

$$\begin{align*} &U\land V=U\cap V,\\ &U\lor V=\operatorname{int}_X\,\operatorname{cl}_X(U\cup V),\\ &\lnot U=X\setminus\operatorname{cl}_XU,\text{ and}\\ &U\le V\text{ iff }U\subseteq V\;; \end{align*}$$

it’s well-known that this makes $\mathrm{RO}(X)$ a complete Boolean algebra. Clearly this algebra is atomless if $X$ has no isolated points. In particular, $\mathrm{RO}(\Bbb R)$ is atomless. Since $\Bbb R$ is second countable, it’s clear that $|\mathrm{RO}(\Bbb R)|\le 2^\omega$. On the other hand, for any $A\subseteq\Bbb Z$ the set $$\bigcup_{n\in A}\left(n-\frac14,n+\frac14\right)$$ is regular open, and there are clearly $2^\omega$ such sets, so $|\mathrm{RO}(\Bbb R)|=2^\omega$. In short, $\mathrm{RO}(\Bbb R)$ is a complete, atomless Boolean algebra of power $2^\omega$.

Now let $\mathscr{B}=\wp(\omega)/[\omega]^{<\omega}$, the quotient of the power set algebra of $\omega$ by the ideal of finite subsets of $\omega$. Since $|\wp(\omega)|=2^\omega$, and $\left|[\omega]^{<\omega}\right|=\omega$, it’s clear that $|\mathscr{B}|=2^\omega$. It’s also clear that $\mathscr{B}$ is atomless. However, $\mathscr{B}$ is not complete, so it most be distinct from $\mathrm{RO}(\Bbb R)$.

To see that $\mathscr{B}$ is not complete, let $\{A_n:n\in\omega\}$ be a partition of $\omega$ into infinite subsets. For each $A\subseteq\omega$ denote by $\widehat A$ its equivalence class in $\mathscr{B}$. If $\widehat S$ is any upper bound in $\mathscr{B}$ for $\{\widehat{A_n}:n\in\omega\}$, $|A_n\setminus S|<\omega$ for each $n\in\omega$, so for each $n\in\omega$ we may choose $s_n\in S\cap A_n$. Let $T=\{s_n:n\in\omega\}$, and let $S\,'=S\setminus T$; clearly $|A_n\setminus S\,'|<\omega$ for each $n\in\omega$, so $\widehat{S\,'}$ is an upper bound for $\{\widehat{A_n}:n\in\omega\}$, but it’s also clear that $\widehat{S\,'}<_{\mathscr{B}}\widehat{S}$.

(One could also note that $\mathrm{RO}(\Bbb R)$ is ccc, while $\mathscr{B}$ isn’t, but t.b. already used that idea, so I thought that I’d do something different.)

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Let $B_0$ be the free Boolean algebra generated by $2^{\aleph_0}$ generators. Observe that it has $2^{\aleph_0}$ elements and by construction descending chains of length $\omega$. On the other hand take the language of boolean algebras and add $2^{\aleph_0}$ many constants, $(c_\alpha)_{\alpha<2^{\aleph_0}}$. Take the theory that contains the theory of atomless Boolean algebras plus the sentences $c_\alpha<c_\beta$ for all $\alpha>\beta$. This is certainly finitely satisfiable by the countable atomless Boolean algebra. Hence by compactness it is satisfiable and it has a model $B_1$. In fact by Lowenheim-Skolem theorem you can make $B_1$ of size $2^{\aleph_0}$.

Now $B_0$ and $B_1$ are not isomorphic since one has descending sequences of size $\omega$ only while the other contains a descending sequence of size $2^{\aleph_0}$.

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I'll try:

If I'm not mistaken, the cardinality of the completion of the countably infinite atomless Boolean algebra is that of the continuum.

For the second example, take the Boolean algebra freely generated by the set of all reals.

But I'm not altogether sure those are not isomorphic, so this answer is not fully complete.

Later edit: Apostolos has now posted my second example plus another, which he shows is not isomorphic to it.

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