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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?

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2 Answers 2

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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.

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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!

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  • $\begingroup$ Oh I see. You well-order the set and then in each element you scrape out the element-elements that already appeared earlier. $\endgroup$ Dec 6, 2023 at 2:36
  • $\begingroup$ @GregNisbet Yup. But without choice this doesn't work. $\endgroup$ Dec 6, 2023 at 2:40
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    $\begingroup$ 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. $\endgroup$
    – bof
    Dec 6, 2023 at 3:18
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    $\begingroup$ 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 $g : P(X) \setminus \{ \emptyset \} \to X$. $\endgroup$ Dec 6, 2023 at 17:39
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    $\begingroup$ @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. $\endgroup$
    – user21820
    Dec 7, 2023 at 4:44

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