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Definition 1. For all collections $\mathcal{C}$ whose elements are sets, and cardinal numbers $\kappa$, define that $\mathcal{C}$ is closed under $\kappa$-limited unions iff for all $\mathcal{B} \subseteq \mathcal{C}$ such that $|\mathcal{B}|<\kappa$, it holds that $\bigcup \mathcal{B} \in \mathcal{C}$.

Definition 2. For all collections $\mathcal{C}$ whose elements are sets, define that $\mathcal{C}$ is closed under aligned unions iff for all $\mathcal{B} \subseteq \mathcal{C}$ such that (for all $A,B \in \mathcal{B}$ it holds that either $A \subseteq B$ or $A \supseteq B$), it holds that $\bigcup \mathcal{B} \in \mathcal{C}$.

Now suppose $\mathcal{C}$ is closed under $\kappa$-limited unions, and also under aligned unions. What is the greatest cardinal number $\kappa'$ such that we can conclude that $\mathcal{C}$ is closed under $\kappa'$-limited unions?

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Under these hypotheses (and the additonal assumption that $\kappa > 2$) it will follow that $\mathcal{C}$ is closed under arbitrary unions.

We show by induction on $\lambda$ that $\mathcal{C}$ is closed under $\lambda$-limited unions. Note that for $\lambda < \kappa$ the result is true by hypothesis. Suppose that $\lambda > \kappa$ is such that $\mathcal{C}$ is closed under $\mu$-limited unions for all $\mu < \lambda$.

Given $\mathcal{B} \subseteq \mathcal{C}$ with $| \mathcal{B} | = \mu < \lambda$ we may enumerate $\mathcal{B}$ as $\{ B_\alpha : \alpha < \mu \}$. For each $\alpha < \mu$ define $A_{\alpha} = \bigcup_{\xi \leq \alpha} B_\xi$. As $\mathcal{C}$ is closed under $\mu$-limited unions it follows that $A_\alpha \in \mathcal{C}$ for all $\alpha < \mu$, and since $\mathcal{C}$ is closed under aligned unions it follows that $\bigcup_{\alpha < \mu} B_\alpha = \bigcup_{\alpha < \mu} A_\alpha \in \mathcal{C}$.


To see that some bound on $\kappa $ is needed, consider the family of all singletons in some set of cardinality $> 1 $.

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I think somewhere you need to assume that $\kappa \geq 2$. –  goblin Mar 5 '13 at 14:34
2  
@user18921: Aren't all cardinals infinite? ;-) –  Arthur Fischer Mar 5 '13 at 14:38

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