TITLE: Can we prove the existence of $A\cup B$ without the union axiom?
QUESTION [23 upvotes]: If $A$ and $B$ are sets, then $A\cup B$ is defined as follows:
$$A\cup B:=\{x: x\in A \, \, \,\text{ or } \, \, \, x\in B\}.$$
In ZFC, $A\cup B$ exists because $\bigcup\{A, B\}$ exists by union axiom, and it is known that axiom of union is independent of rest of ZFC.
If there is a set such that $A\subseteq C$ and $B\subseteq C$, then we can define $A\cup B$ using the $C$ and separation axiom. But can I prove the existence of $C$ in ZF-Union? Thanks for any help.

REPLY [9 votes]: As Andrés' answer points out, ZFC - Union proves the existence of finite unions, and much more. AC is in fact essential for this theorem, for ZF proves Con(ZF - Union + "there are $x,y$ such that $x \cup y$ doesn't exist").
Recall that $\aleph(x)$ is defined to be the least ordinal which does not inject into $x$ and $\aleph^*(x)$ the least ordinal which $x$ does not surject onto.
Lemma: For infinite $A,$ define $H'(A)= \{x: \forall y \in TC(\{x\}) (|y| \le |A|) \}.$ Then $H'(A) \subset V_{\aleph^*(A)^+}.$ In particular, $H'(A)$ is a transitive set.
Proof of lemma: Let $\kappa=\aleph^*(A).$ It suffices to find a surjection from $\kappa$ onto $\text{rk}``(TC(\{x\})$ for $x \in H'(A).$
We will recursively define surjections $f_{x,n}: \kappa \rightarrow \text{rk}``(\bigcup^n \{x\})$ for all $x \in H'(A).$ Suppose we have defined all such functions up through $n.$ Define a surjection $g_{x, n+1}: \kappa \times x \rightarrow \text{rk}``(\bigcup^{n+1} \{x\})$ by $(\alpha, y) \mapsto f_{y,n}(\alpha).$ For each $\alpha, \text{ot}(g_{x, n+1}``(\{\alpha\} \times x))<\kappa,$ yielding a canonical surjection $f_{x, n+1}: \kappa \simeq \kappa \times \kappa \rightarrow \text{rk}``(\bigcup^{n+1} \{x\}).$ This completes the recursion, and now we define a surjection $f_x: \kappa \simeq \kappa \times \omega \rightarrow \text{rk}``(TC(\{x\}))$ by $(\alpha, n) \mapsto f_{x, n}(\alpha).$
Now, to prove the main result, we work in the Bristol model, a certain submodel of $L[c]$ ($c$ a Cohen real over $L$) which, among many interesting properties, satisfies ZF + existence of a set $A$ such that $A$ does not inject into $\mathcal{P}^n(\alpha)$ for any ordinal $\alpha$ and $n<\omega.$ (See https://arxiv.org/pdf/1704.06939.pdf) Fix such an $A.$ Let $\kappa = \aleph(\mathcal{P}^{\omega}(A)).$
Let $M=\{x: \forall y \in TC(\{x\}) \exists n<\omega(|y| \le |\mathcal{P}^n(A)| \vee |y| \le |\mathcal{P}^n(\kappa)|)\}.$ Applying the lemma to $\mathcal{P}^{\omega}(A \cup \kappa),$ we see that $M$ is a transitive set. We check that $M \models \text{ ZF - Union + there are $x,y$ such that $x \cup y$ doesn't exist}).$
We get extensionality and foundation from $M$ being transitive. Pairing, infinity, and power set are clear. We get separation and replacement from $M$ being closed under subset and surjection  (noting that if $x$ injects into $y,$ any surjective image of $x$ injects into $\mathcal{P}(y)$). Finally, $A, \kappa \in M,$ yet $A \cup \kappa \not \in M.$