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Is the following conjecture about collections of sets correct? And if so, does the proof work?

Conjecture. Let $\kappa$ denote an infinite cardinal number, and let $\mathcal{K}$ denote a collection of sets that is closed with respect to the following.

  1. Disjoint unions of cardinality less than or equal to $\kappa$.
  2. Complements (with respect to some set that is larger than or equal to $\bigcup \mathcal{K}$).
  3. Unions and/or intersections of cardinality strictly less than $\kappa$ (not necessarily disjoint).

Then $\mathcal{K}$ is closed with respect to unions and intersections of cardinality less than or equal to $\kappa$ (not necessarily disjoint).

Proof. Let $\mathcal{J} \subseteq \mathcal{K}$ denote a subcollection of cardinality less than or equal to $\kappa$. It will be shown that $\bigcup \mathcal{J} \in \mathcal{K}$.

Let $\lambda$ denote the least ordinal such that $|\lambda| = \kappa$, and choose a bijection $j : \lambda \rightarrow \mathcal{J}$. Define a function $k : \lambda \rightarrow \mathcal{K}$ as follows.

$$k(\beta) = \left( \bigcup_{\alpha<\beta} j(\alpha)\right)^c \cap j(\beta).$$

We will show that for all $\beta \in \lambda$ it holds that $k(\beta) \in \mathcal{K}$.

Fix $\beta \in \lambda$. Since $\lambda$ is the least ordinal with cardinality $\kappa$ and $\beta \in \lambda$, the union in the above expression has cardinality strictly less than $\kappa$, and is thus an element of $\mathcal{K}$. Thus its complement is an element of $\mathcal{\kappa}$. But since $\kappa$ is infinite, it follows that $\mathcal{K}$ is closed with respect to finite intersections, thus the above expression is an element of $\mathcal{K}$. So $k(\beta) \in$ $\mathcal{K}$.

We conclude that for all $\beta \in \lambda$ it holds that $k(\beta) \in \mathcal{K}$. Furthermore, it can be shown that $$\bigcup_{\beta \in \lambda}k(\beta) = \bigcup_{\beta \in \lambda}j(\beta).$$

Noting that the expression on the left is a disjoint union of elements of $\mathcal{K}$ and has cardinality $\kappa$, we conclude that it is an element of $\mathcal{K}$. Thus the expression on the right is an element of $\mathcal{K}$. But the expression on the right equals $\bigcup \mathcal{J}.$ Therefore, $\bigcup \mathcal{J} \in \mathcal{K}$, as required.

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  • $\begingroup$ Sounds sound. You don't always get a bijection $j$ (later called $g$?) if $\mathcal J$ is too small - but of course then (3) already shows that there is nothing to show. Next, I'm always worried when I read complements, so let us assume that $\bigcup \mathcal K$ is a set or restrict the notion of complements to (the what your proof really uses, namely) relative complements to a set in $\mathcal K$ ("collection of sets" seemd to indicate that you'd allow $\mathcal K$ to be a class). $\endgroup$ – Hagen von Eitzen Feb 23 '13 at 9:32
  • $\begingroup$ Yeah by complements I meant "complements in some larger, unspecified set." Also, I fixed the $j$/$g$ issue (thanks!). EDIT: Why do you say that you don't always get a bijection? I think that if $\mathcal{J}$ is empty, then the empty bijection does the trick. $\endgroup$ – goblin Feb 23 '13 at 11:05
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I think the conditions can be made stronger:

Theorem. Assume the class of sets $\mathcal K$ has these properties, where $\kappa$ is any cardinal:

  1. If $|I|\le \kappa$ and $A_i\in \mathcal K$ for $i\in I$ and $A_i\cap A_j=\emptyset$ if $i\ne j$, then $\bigcup_{i\in I}A_i\in\mathcal K$.
  2. If $A,B\in\mathcal K$ then $A\setminus B\in \mathcal K$.

Claim: If $|I|\le \kappa$ and $A_i\in \mathcal K$ for $i\in I$, then $\bigcup_{i\in I}A_i\in\mathcal K$.

Proof: Let $$\mathcal L=\biggl\{\beta\in\mathrm{On}\biggm| |\beta|\le \kappa\text{ and }\exists f\colon\beta\to\mathcal K\text{ such that }\bigcup_{\alpha<\beta}f(\alpha)\notin\mathcal K\biggr\}.$$

Assume $\mathcal L\ne\emptyset$ and let $\lambda=\min\mathcal L$. Fix some $j\colon\lambda\to\mathcal K$ with $$S:=\bigcup_{\alpha<\lambda}j(\alpha)\notin\mathcal K.$$ For $s\in S$ let $i(s)=\min\{\alpha\in\lambda\mid s\in j(\alpha)\}$. Then for $\alpha\in\lambda$, $$ i^{-1}(\alpha)=j(\alpha)\setminus\bigcup_{\beta<\alpha}j(\beta).$$ Since $\alpha<\lambda$, we have $\alpha\notin\mathcal L$ and $|\alpha|\le\kappa$, hence $\bigcup_{\beta<\alpha}j(\beta)\in\mathcal K$ and hence by property (2) also $i^{-1}(\alpha)\in\mathcal K$. Now $$ S=\dot\bigcup_{\alpha\in\lambda}i^{-1}(\alpha)$$ shows that $S$ can be written as a disjoint union of $|\lambda|$ elements of $\mathcal K$. Since $|\lambda|\le\kappa$ and $S\notin \mathcal K$, we obtain a contradiction with property (1). Therefore $\mathcal L=\emptyset$.

Final conclusion: If $I$ with $|I|\le\kappa$ is given, then there exists an ordinal $\beta$ and a bijection $g\colon\beta\to I$. With $f\colon\beta\to\mathcal K$, $\alpha\mapsto A_{g(\alpha)}$ we find that $$\bigcup_{i\in I} A_i=\bigcup_{\alpha<\beta}f(\alpha)\in\mathcal K$$ because $|\beta|=|I|\le\kappa$ and $\mathcal L=\emptyset$. $_\square$

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  • $\begingroup$ "I think the conditions can be made stronger." Do you mean weaker? Anyway thanks for the reply, I'll check your proof out tomorrow, when I've had some sleep. $\endgroup$ – goblin Feb 23 '13 at 11:08

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