About the axiom of union, from the naive set theory is very natural understand the concept of union (as well others Boolean operation) between the subset os a fixed base set X, this is because $X$ is like a classification of its own elements, then is thinkable for each element $x$ and subset $A$ of $X$ if $x$ is in $A$ or not, this is because $X$ as collection fix a "determinism" (or a well founded conceptual control) on its elements and on its subsets. But is $x$ and $Y$ are different subset (totally different as context) what could be a criterion for think that a set $x$ (view as element) belong to $Y$?. Set theory is strong just because forget all possible questions or consideration about type or nature of elements, these are merely abstract, then how is thinkable that a general $x$ is in a general set $Y$? Axiom of extensionality seems give a formal and (ideally) operative answer to this: $x$ is in $Y$ (write $x\in Y$) iff exist a $y$ in $Y$ such that $x=y$ but this last is true iff (for the some axiom) elements of $x$ are equal to elements of $y$ and viceversa, and this could come a endless iteration. I know that these questions or nonsense for a pure formalitic point of view, Set theory is merely a set of rule and axiom of a precise first order logic theory, and $x\in Y$ dont mean nothing, but a formal formula. But when you start to make mathematics (or think about) do make a foundations, you haven't the mathematical logics, it is inside the mathematics too, you have the classical general Aristotle logic, and set theory before becoming the background for building clay for make any mathematical structure (the models of itself included) start from considerations that are half philosophical too.

But stop to these surreal considerations.

I ask, is we make a set theory $ZF'$ replacing the union axiom by the more weak:

WU) Give a family of subset $U_i\ i\in I$ ($I$ a set) of some set $X$, then exist the union

$\cup_{i\in I}U_i=${ $x\in X|\exists i\in I: x\in U_i $}.

(i.e. localize the union axiom, as the separation axiom is the localization of axiom of formation of classes for avoid the RUssel paradox)

and keep all other $ZF$ axioms, what could be the limitation of this new theory $ZF'$ respect to $ZF$?

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    $\begingroup$ The axiom WU is an instance of the Comprehension Axiom, so ZF' is just ZF without the Union Axiom. I believe that ZF' cannot prove that the union of two sets always exists, but that might depend on how the other axioms are formulated. Could you give a reference version for the axioms of ZF' (especially the Replacement Axiom)? $\endgroup$ – François G. Dorais Nov 21 '11 at 18:09
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    $\begingroup$ ZF' + “unions of two sets exist” is still much weaker than ZF (in the sense that ZF proves its consistency: see mathoverflow.net/questions/48365/…). $\endgroup$ – Emil Jeřábek Nov 21 '11 at 18:20
  • $\begingroup$ I dont know if exist a reference for ZF', i think to the WU because the general set union is too bad for have a categorical traduction, in the weak form WU is easly definible in terms (internal logic) of topos theory. Anyway I seem that Comprehension axiom allow the union of "definible" (by some formulas) subset. Thank for your interest anyway. $\endgroup$ – Buschi Sergio Nov 21 '11 at 18:37
  • $\begingroup$ When interpreted in category-theoretic terms, the Union Axiom is mostly there to ensure that Set (viewed as a topos) is closed under internal coproducts. This is true in any elementary topos: if $f:A\to B$ is any morphism, then $f^*$ has both a left adjoint (internal coproduct) and a right adjoint (internal product). $\endgroup$ – François G. Dorais Nov 21 '11 at 18:59
  • $\begingroup$ I believe ZF' proves that the union of two sets always exists (at least, on what I think is a natural interpretation of what ZF' is to be): Given sets $A$ and $B$, ZF' can establish the existence of $P(A) \times P(B)$ [powersets being given directly by an axiom and cartesian products requiring only the axiom of pairing (on top of extensionality)]. From this, we can use Separation to extract the subset of such pairs with one component an empty set and the other a singleton (in either A or B); finally, we can apply replacement to this to obtain the union of A and B. $\endgroup$ – Sridhar Ramesh Nov 23 '11 at 11:17

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