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Suppose $\mathfrak{A}$ is a Boolean algebra and $\mathfrak{J}$ is chain of ideals in $\mathfrak{A}$ ordered by inclusion such that none of its elements is countably generated. Clearly, the union $\bigcup \mathfrak{J}$ is an ideal. Can it be countably generated?

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Take te interval algebra, $B$, of the ordinal $\omega_1\times\omega$; for $n\in\omega$ let $I_n$ be the ideal consisting of the elements of $B$ that are bounded below $\omega_1\times n$. None of the ideals $I_n$ is countably generated but the union $\bigcup_n I_n$ is countably generated, by the family $\lbrace \omega_1\times n:n\in\omega\rbrace$. I assume here that you mean that $A$ generates $I$ in case $A\subseteq I$ and for all $x\in I$ there is $a\in A$ such that $x\le a$.

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Thank you. For me, an ideal $I$ is generated by $A$ if $I$ is the intersection of all ideals containing $A$. Your definition is clearly equivalent to mine for $A=\{p\}$ but I am not sure if this actually the same for infinite generating sets. –  GiroCont Nov 23 '11 at 12:44
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GiroCont, you need only take the collection of finite joins of elements of your generating set to get a generating set in the sense of KP Hart. But the set of finite joins from a countable set is still countable, and so the two notions for the ideal are equivalent. –  Joel David Hamkins Nov 23 '11 at 14:20
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Let $\mathfrak{A}$ be the power set of an uncountable set $X$, which is a complete Boolean algebra. Select disjoint sets $X_n\subset X$ of size $\omega_1$, and let $J_n$ be the ideal generated by $X_0\sqcup\cdots\sqcup X_n$, together with the countable subsets of $X_{n+1}$. Thus, each ideal $J_n$ is countably complete (closed under countable unions) and not countably generated, and they form a chain $J_0\subset J_1\subset\cdots$. The union $\bigcup_n J_n$, however, is simply the ideal generated by the sets $X_n$.

The general situation of this example (and also of KP Hart's example, which is fundamentally similar) is to have a tower of ideals $J_n$, such that nested between them are principal ideals $I_n$, so that the chain looks like: $$J_0\subset I_0\subset J_1\subset I_1\subset\cdots$$ The point is that the union of the ideals $J_n$ is the same as the union of the principal ideals $I_n$, which is of course countably generated because they are principal.

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It would suffice if the interleaving ideals $I_n$ were countably generated, rather than principal. Thus, any chain of non-countably generated ideals $J_n$ for which we can find countably generated interleaving ideals $I_n$ will serve as an example. –  Joel David Hamkins Nov 23 '11 at 14:22
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