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.

  • $\begingroup$ 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

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 $\mathcal L=\emptyset$. $_\square$

  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.