10

The unique union $\mathcal{U}(F)$ is defined as $\{x \mid [\exists! A \in F](x \in A)\}$.

I saw this question earlier today, and I was wondering what one might reasonably use the unique union construction for.

I think a natural first question to ask is whether ZFC -Union +UniqueUnion is equivalent to ZFC.

Let $A \oplus B$ be $\mathcal{U}(\{A, B\})$, which exists by pairing.

We can define the binary union $A \cup B$ using $A \oplus B \oplus A\cap B$, noting that $A \cap B$ is $\{x \mid x \in A \land x \in B \}$, which exists by comprehension.

By a result quoted in this answer (which I do not understand at all), it is consistent with ZFC -Union that there exist two sets $x$ and $y$ whose union does not exist. Although, by a result quoted in this other answer to the same question, it cannot be the case that $x$ and $y$ are both finite.

ZFC -Union +UniqueUnion does rule out the possibility of two sets whose union doesn't exist, so it is stronger than ZFC -Union. This makes sense. It seems intuitively reasonable that the existence of the unique union is not a theorem of the other axioms.

How do ZFC and ZFC -Union +UniqueUnion compare, though?

Greg Nisbet
  • 11,657

2 Answers2

16

I think the equivalence can be proved using Replacement and Power Set instead of Choice.

  1. Binary Union: given sets $A$ and $B$ we get $A\cup B=\mathcal U\{A,B\setminus A\}$ by using Specification, Pairing, and Unique Union.

  2. Given a set $A$ we get $\{A\}\times A\subseteq\mathcal P\mathcal P(A\cup\mathcal PA)$ by Power Set, Binary Union, and Specification.

  3. Given a set $\mathcal A$ we get $\{\{A\}\times A:A\in\mathcal A\}$ by Replacement, and then

  4. we get $\{\langle A,a\rangle:a\in A\in\mathcal A\}=\mathcal U\{\{A\}\times A:A\in\mathcal A\}$ by Unique Union,

  5. and finally we get $\bigcup A$ from $\{\langle A,a\rangle:a\in A\in\mathcal A\}$ by Replacement.

bof
  • 78,265
5

They are equivalent.

Suppose $X$ is a set whose union we want to take. Well-order $X$ to get $X=(x_\alpha)_{\alpha<\kappa}$ for some $\kappa$. By $\mathsf{Replacement}$ we can define $\mathscr{Y}=(y_\alpha)_{\alpha<\kappa}$ such that $$y_\alpha=\{z\in x_\alpha: \forall \beta<\alpha(z\not\in x_\beta)\}.$$ Now apply unique union to (the range of) $\mathscr{Y}$.

Interestingly, I don't see a way to do without choice!

Noah Schweber
  • 245,398
  • Oh I see. You well-order the set and then in each element you scrape out the element-elements that already appeared earlier. – Greg Nisbet Dec 06 '23 at 02:36
  • @GregNisbet Yup. But without choice this doesn't work. – Noah Schweber Dec 06 '23 at 02:40
  • Towards getting rid of choice, note that we only need to overshoot: with $\mathsf{Separation}$, we can always pare things down later. But I don't see a way to do that. – Noah Schweber Dec 06 '23 at 02:41
  • 2
    Well-order $X$ to get $X=(x_\alpha)_{\alpha\lt\kappa}$ for some $\kappa$? So the Axiom of Union was never used in the development of ordinal numbers and the proof of the well-ordering theorem? I'm not saying it was, I wouldn't know, you're the expert. It just seems to me like that's a long way to go without using a basic axiom. – bof Dec 06 '23 at 03:18
  • 2
    The proof of $AC \implies WO$ I'm familiar with makes essential use of the union axiom, in taking the union of all well-orders of a subset $A \subseteq X$ satisfying $x = g(A \setminus { y\in A \mid y < x })$ for some choice function $g : P(X) \setminus { \emptyset } \to X$. – Daniel Schepler Dec 06 '23 at 17:39
  • 1
    @DanielSchepler: That's not an essential issue. Let C∈ChoiceFunc(Pow(X)∖{∅}). Let V = { ⟨t,u⟩ : t,u∈X ∧ ∃S⊆X ∃◁∈WO(S) ( t ◁ u ∧ ∀i∈S ( i = C(S∖S[◁i]) ) ) }. You don't have to collect all the towers before taking their union, since you can just collect the individual pairs. – user21820 Dec 07 '23 at 04:44