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Suppose the set theory we are working with implies that $\forall x \forall y \exists(x \cup y)$.

Would that implication alone further imply the existence of an arbitrary union?:

$$\forall S\exists(\bigcup_{x\in S}S)$$

As such could be defined as:

$$\bigcup_{x\in S}S=x_1\cup x_2\cup x_3\cup \dots \forall x_i\in S$$

This is trivially true for finite sets. However, would infinite sets complicate the matter?

I should also point out that I understand that most set theories (such as ZF) first define the arbitrary union (perhaps even to avoid this question), which would retroactively account for binary unions. For the purpose of this question, assume that the arbitrary union is not directly implied within the axiomatic system we're working with, meanwhile the binary union is.

Graviton
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    The existence of finite unions does not imply that of infinite unions. For instance, $H(\beth_\omega)$, the collection of sets of hereditary cardinality smaller than $\beth_\omega$, is a model of ZFC - union, and in this model finite unions exist, but not countable ones. – Andrés E. Caicedo Jun 04 '20 at 04:44
  • @AndrésE.Caicedo That seems intuitive. Although, I'm unfamiliar with $H$ and $\beth$, would you direct me to where I can learn about such? – Graviton Jun 04 '20 at 05:05
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    Jech's or Kunen's set theory textbooks cover these notions. Either one would be an excellent reference. – Andrés E. Caicedo Jun 04 '20 at 05:57
  • @AndrésE.Caicedo Why does $H(\beth_\omega)$ not satisfy Union? The transitive closure of the union is contained in the transitive closure of the original set, or am I missing something? It does not satisfy Replacement though. – Jonathan Schilhan Jun 06 '20 at 16:02

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