# Does ZFC -Union +UniqueUnion prove the axiom of union?

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?

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.

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!

• Oh I see. You well-order the set and then in each element you scrape out the element-elements that already appeared earlier. Dec 6, 2023 at 2:36
• @GregNisbet Yup. But without choice this doesn't work. Dec 6, 2023 at 2:40
• 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 6, 2023 at 3:18
• 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 $g : P(X) \setminus \{ \emptyset \} \to X$. Dec 6, 2023 at 17:39
• @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. Dec 7, 2023 at 4:44