There are some questions posted on the countable case (Countable Union to Countable Disjoint Union; Writing an Infinite Union as the Union of Disjoint Sets; union of sets disjoint sets is equal to the union set difference). I can't see why this wouldn't extend to the uncountable case, and would appreciate correction or confirmation for my proof below......

Proposition: if $\{B_α\}_{ α ∊ A }$ is any collection of sets and $B = \cup_{ α ∊ A} B_α$ then there is a collection of pairwise disjoint sets $\{C_α\}_{ α ∊ A }$ with $B = \cup_{ α ∊ A} C_α$. Furthermore, for all $α, C_α \subset B_α$ (So, this applies to uncountable, countably infinite, and finite collections).

Proof: By the axiom of choice we can assume that $A$ is well-ordered with a minimum element $α_0$, and that every non-empty subset of $A$ has a minimum element (which is relevant in steps (2) and (3)).
Define $C_{α_0} = B_{α_0}$ and for $α > α_0$, $C_α = B_α \setminus (\cup_{ β < α} B_β)$ .

  1. Then for all $α, C_α ⊂ B_α$.
  2. Among sets {$C_α$} which contain some particular element $x$, let $γ$ be the smallest value of $α$. Then $x ∊ C_γ ⇒ x ∊ B_γ$ and $x ∉ B_β$ for $β < γ$. So $x ∉ C_β$ for $β < γ$.
    And, $x ∊ \cup_{ β < α} B_β$ for $α \ge γ ⇒ x ∉ C_α$ for $α > γ$.
    Therefore the sets {$C_α\}_ {α ∊ A}$ are pairwise disjoint.
  3. Now for some particular element $x \in B = \cup_{ α ∊ A} B_α$, let $γ $ be the smallest value of $α$ such that $x \in B_γ$. Then $x ∊ C_γ ⇒ x \in \cup_{ α ∊ A} C_α ⇒ B = \cup_{ α ∊ A} B_α \subset \cup_{ α ∊ A} C_α$.
    Conversely, for some particular element $x \in \cup_{ α ∊ A} C_α$ , let $γ$ be the smallest value of $α$ such that $x ∊ C_γ$. Then $x \in C_γ ⇒ x \in B_γ ⇒ \cup_{ α ∊ A} C_α \subset \cup_{ α ∊ A} B_α$.
    So, $B = \cup_{ α ∊ A} B_α = \cup_{ α ∊ A} C_α$.
  • $\begingroup$ Right on. Moreover, I think your Proposition is equivalent to the axiom of choice. $\endgroup$ – bof Aug 24 '17 at 11:02
  • $\begingroup$ By "every subset of $A$ has a minimum element" you mean (of course) "every nonempty subset of $A$ has a minimum element." But you don't have to say that, that's what it means for $A$ to be well-ordered. $\endgroup$ – bof Aug 24 '17 at 11:52
  • $\begingroup$ @bof Yes, non-empty. I haven't seen how to prove the reverse for equivalence - want to post it as an answer ? $\endgroup$ – Tom Collinge Aug 24 '17 at 13:48

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