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Short version: Is the axiom of union independent of the rest of axioms of ZF?

NO) Tourlakis (2003) says in p. 177 that the axiom of union can be derived from the rest of ZF if an appropriate version of collection axiom[*] is chosen. The quote is:

«Bourbaki (1966b) adopts the axiom of pairing, but adopts collection version (2), and proves both separation and union»

YES) In the other hand, I have read here and here something like "$H_{\kappa}$ is a model for ZF-Union+¬Union", where $\kappa$ was $\beth_\omega$ or a singular cardinal.

Any reference on the subject would be highly appreciated. I apologize in advance if the question is too basic (not a mathematician!). Also, I have googled it and followed some false trails before asking here. Thanks.

[*] The appropriate version of collection is apparently weaker (or equivalent at most) than the collection axiom that he is adopting in his text. I think the statement is:

$(∀x)(∃z)(∀y)({\mathcal P} [x, y] → y ∈ z) → (\forall A)(\exists B)(\forall y)(y\in B\leftrightarrow (\exists x\in A){\mathcal P}[x,y])$

I have translated the notation from III.8.12 and III.2 (obviating any reference to ur-elements).

EDIT: Thank you very much for the answers, they were really helpful.

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  • $\begingroup$ Can you write down what the "appropiate version of collection" is? $\endgroup$
    – Ramiro de la Vega
    Commented Nov 24, 2011 at 17:47
  • $\begingroup$ If the answers ware helpful, pick one that helped the most and 'accept' it by clicking on the tick. This is better than editing your question. $\endgroup$
    – David Roberts ♦
    Commented Nov 24, 2011 at 23:10

3 Answers 3

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The usual version of collection is the second one in Kaveh's answer. It, with the remaining axioms, won't give the axiom of union. A counterexample is given in the question, except for a slightly unusual definition of $H_\kappa$ for singular $\kappa$; it should be the collection of those sets $x$ such that each member of the transitive closure $TC(\{x\})$ has cardinality $<\kappa$. (This does not imply that the whole $TC(\{x\})$ has cardinality $<\kappa$, which is the usual meaning of $H_\kappa$.)

The "appropriate version" of collection used by Bourbaki seems appropriate mainly in the sense that the axiom of union has been built in.

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  • $\begingroup$ The claim about the "weakness" of the "appropriate" version of the collection axiom still puzzled me; but I realize now that it is probably irrelevant, since the implication is proved using the rest of axioms of ZF (including union). $\endgroup$
    – 0rvidal
    Commented Nov 24, 2011 at 23:05
  • $\begingroup$ Shoenfield, Mathematical Logic also uses the "appropriate" version. Do I understand correctly that you are saying there are reasons to prefer the "usual" version? $\endgroup$
    – rgrig
    Commented Nov 25, 2011 at 13:08
  • $\begingroup$ As far as I can see, I think there is no special reason to adopt the "appropriate" version of Bourbarki or Shoenfield (but remember: I am not an specialist, not even a mathematician!). I simply got confused about of the independence of Unions by the statements in Tourlakis (2003). $\endgroup$
    – 0rvidal
    Commented Nov 25, 2011 at 14:14
  • $\begingroup$ @Tct: Well, I'm asking if there is a reason to choose the usual one, which is not the same as asking if there is a reason not to choose the appropriate one. :P $\endgroup$
    – rgrig
    Commented Nov 25, 2011 at 17:06
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The axiom is

$$\forall x \ \exists y \ \forall z \ (\varphi(z,x) \to z\in y) \to \forall X \ \exists Y \ \forall y \ (y \in Y \leftrightarrow \exists x\in X \ \varphi(y,x)) $$

Consider

$$\forall x \ \exists y \ \forall z \ (z\in x \to z\in y) \to \forall X \ \exists Y \ \forall y \ (y \in Y \leftrightarrow \exists x\in X \ y\in x ) $$

This is an instance of the axiom. Now the left side is true (take $y=x$). The right side expresses the existence of the union of $X$.

The other version of the axiom:

$$\forall x\in X \ \exists y \ \varphi(x,y) \to \ \exists Y \ \forall x\in X \ \exists y\in Y \ \varphi(x,y)$$

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    $\begingroup$ The second part of your answer doesn't seem to do what you want, since $Y$ could just be the given $X$, not its union (assuming $P$ is intended to mean power set). $\endgroup$
    – Andreas Blass
    Commented Nov 24, 2011 at 19:26
  • $\begingroup$ @Andreas, thank you. $\endgroup$
    – Kaveh
    Commented Nov 24, 2011 at 19:37
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Let me add a nice remark I just became aware of: In

Greg Oman. On the axiom of union, Arch. Math. Logic, 49 (3), (2010), 283–289. MR2609983 (2011g:03122),

Oman clarifies precisely which unions can be proved to exist in $\mathsf{ZFC}-\mathrm{Union}$: $\bigcup x$ exists iff $\{|y|\colon y\in x\}$ is bounded. In particular, $A\cup B$ exists for any sets $A,B$.

See this MSE question for details. The proof uses in essential ways both choice and replacement.

(Curiously, I do not know of an original reference for the fact that $\mathsf{ZFC}-\mathrm{Union}$ does not suffice to prove the existence of infinite unions, the usual argument being the one in the body of the question, and in Andreas's answer. It would be nice to have the reference, so it can be added here.)

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  • $\begingroup$ When you say bounded, do you mean...? $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 19, 2013 at 22:02
  • $\begingroup$ There is an ordinal $\alpha$ such that $|y|\le\alpha$ for all $y\in x$. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 19, 2013 at 22:12
  • $\begingroup$ But... isn't that true for any set in $\sf ZFC$? $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 19, 2013 at 22:16
  • $\begingroup$ Yes. We are working in $\mathsf{ZFC}-\mathrm{Union}$. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 19, 2013 at 22:18
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    $\begingroup$ I'm sure that if I would have had that extra Guinness I would have picked up on the fact that there might be a cofinal set of ordinals... :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 20, 2013 at 8:12

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