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I have the following problem:

Given $\{A_{\lambda}\}_{\lambda \in \Lambda}$ for some index set $\Lambda$. Prove that there exists $\{B_{\lambda}\}_{\lambda \in \Lambda}$ such that:

1) $(\forall \lambda \in \Lambda)(B_{\lambda} \subseteq A_{\lambda})$

2) $(\forall \lambda_1 \in \Lambda)(\forall \lambda_2 \in \Lambda)(\lambda_1 \neq \lambda_2 \Rightarrow B_{\lambda_1} \cap B_{\lambda_2} = \emptyset)$

3) $\bigcup_{\lambda \in \Lambda}A_{\lambda} = \bigcup_{\lambda \in \Lambda}B_{\lambda}$

My idea is to define the set of those $\{B_{\lambda}\}_{\lambda \in \Lambda}$ for which properties 1 and 3 are true, partially order it by $\{B'_{\lambda}\}_{\lambda \in \Lambda} \leq \{B''_{\lambda}\}_{\lambda \in \Lambda} \Leftrightarrow (\forall \lambda \in \Lambda)(B'_{\lambda} \subseteq B''_{\lambda})$ and apply Zorn's lemma to this set. However, I'm not sure this idea works. And if it does I'm not sure how to prove that every chain has upper bound. Is this problem-specific or is there something common for most problems? Proving the property of the minimum element also eludes me. Can someone help me?

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If you really want to apply Zorn's lemma directly, here is one possible suggestion. It is similar to your idea, but instead of requiring (1) and (3) we require (1) and (2) to hold. The maximality will guarantee (3).

Define the partially ordered set $(P,\leq)$ where the elements of $P$ are sequences $\{B_\lambda\}_{\lambda\in\Lambda}$, such that for all $\lambda$, $B_\lambda\subseteq A_\lambda$, and the $B_\lambda$'s are pairwise disjoint.

And say that $\{B_\lambda\}\leq\{C_\lambda\}$ if and only if for every $\lambda$, $B_\lambda\subseteq C_\lambda$. That is, pointwise inclusion.

$P$ is not empty because the sequence $B_\lambda=\varnothing$ for all $\lambda$ is there.

Now suppose that you have an increasing chain, take pointwise unions and show that the result must be an element in $P$ (i.e. pairwise disjoint). The deduce from Zorn's lemma that there is a maximal element and from its maximality the union must be equal to the union of the $A_\lambda$.

If you wish to define a partial order in which (1) and (3) hold, then you need to reverse the inclusion, that is "larger" elements in the partial order correspond to small subsets of $A_\lambda$. Then the upper bound would be the pointwise intersection, although it is not immediate to me whether or not (3) must be preserved in this manner. If it does, then again a maximal element should satisfy (2) by its maximality.

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Note that Zorn’s lemma is far from the easiest way to prove this result: you can use the well-ordering principle directly, and the details become both simpler and more natural. Let $\preceq$ be a well-ordering of $\Lambda$, and for each $\lambda\in\Lambda$ let $$B_\lambda=A_\lambda\setminus\bigcup_{\mu\prec\lambda}A_\mu\;.$$ Clearly (1) is satisfied. If $\lambda_1,\lambda_2\in\Lambda$ and $\lambda_1\ne\lambda_2$, then without loss of generality $\lambda_1\prec\lambda_2$, and by construction $B_{\lambda_2}\subseteq A_{\lambda_2}\setminus A_{\lambda_1}\subseteq A_{\lambda_2}\setminus B_{\lambda_1}$, so $B_{\lambda_1}\cap B_{\lambda_2}=\varnothing$, and (2) is satisfied. Finally, let $x\in\bigcup_{\lambda\in\Lambda}A_\lambda$, and let $\mu$ be the $\preceq$-least $\lambda\in\Lambda$ such that $x\in A_\lambda$; then $$x\in A_\mu\setminus\bigcup_{\lambda\prec\mu}A_\lambda=B_\mu\;,$$ so (3) is satisfied as well.

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