1
$\begingroup$

In the Wikipedia article on ZF set theory it says the following about the axiom of union:

Formally, the axiom of union states that for any set of sets $\mathcal{F}$ there is a set $A$ containing every element that is a member of some member of $\mathcal{F}$: $$\forall \mathcal{F} \,\exists A \, \forall Y\, \forall x [(x \in Y \land Y \in \mathcal{F}) \Rightarrow x \in A].$$ While this doesn't directly assert the existence of $\bigcup\mathcal F$, it can be constructed from $A$ in the above using the axiom schema of specification.

Then why use the axiom of union? Is not it redundant?

$\endgroup$
  • $\begingroup$ I can't find any such claim. $\endgroup$ – Asaf Karagila Dec 3 '17 at 14:23
  • $\begingroup$ At: "While this doesn't directly assert the existence of {\displaystyle \cup {\mathcal {F}}} {\displaystyle \cup {\mathcal {F}}}, it can be constructed from {\displaystyle A} A in the above using the axiom schema of specification:" Sorry for not capturing all the formulas. $\endgroup$ – A. R. Dec 3 '17 at 14:25
  • 3
    $\begingroup$ Let this be a lesson: ask accurate and complete questions. Especially if the quoted part is less than three lines. $\endgroup$ – Asaf Karagila Dec 3 '17 at 14:34
6
$\begingroup$

Most axioms of ZF can be phrased in one of two ways:

  1. For every $x$, there is some $y$, such that $y$ is exactly the desired set.

  2. For every $x$, there is some $y$, such that $y$ contains all the elements of the desired set.

These, a priori are not equivalent. But using specification and extensionality, we can actually prove that they are. Because "the desired set" is something which we can define in the language of set theory, so once we have assured there is an even larger set, we can carve it out using specification.

But the second version itself is not somehow provable from specification. You still need to say something about the existence of such a set.

(And this can be applied to union, power set, replacement, and pairing.)


And while we are at it, there's nothing wrong with having a few redundant axioms.

Using the stronger formulation of the axioms (1 in the above part), we can omit specification entirely. Pairing is generally a consequence of a few other axioms. We can include them for simplicity in some cases, or remove them in others.

$\endgroup$
  • $\begingroup$ Thanks for the answer. "You still need to say something about the existence of such a set." Isn't the specification axiom saying something about the existence of sets? The quantifier symbol 'exists' in the specification axiom is right there. A set exists such that it has only those elements (of another set) that have some property and this property could be the one that gives the union. Or this "of another set" does not exist and that is why? "there's nothing wrong with having a few redundant axioms." I think a redundant axiom should not be longer called an axiom. Sorry if I do not understand. $\endgroup$ – A. R. Dec 3 '17 at 14:51
  • $\begingroup$ You posted this as an answer, and it's just a bunch of comments on my question. Let me answer them quickly: (1) no, specification only let's you carve out from existing sets; (2) should the axiom of choice be called an axiom if it can be proved from Zorn's Lemma? (3) ZF is relatively simple and fairly unambiguous. If the only effort you've made to study it is to read the Wikipedia page, the problem does not lie with ZF. It lies with you. $\endgroup$ – Asaf Karagila Dec 3 '17 at 15:07
  • $\begingroup$ I also have some books by Kunen and Jech, but haven't got too far into them. Ok. Specification only let's me carve out from existing sets. So sets do not initially exist. Then why are we talking about them in the axiom of extensionality? I dont understand. In the first axiom we talk about things that do not exist and say that they are unique. And please be open minded about this. $\endgroup$ – A. R. Dec 3 '17 at 15:28
  • $\begingroup$ Extensionality has nothing to do with existence. I suggest you try Lorenz Halbeisen's Combinatorial Set Theory. It has a very good introduction to set theory in its first part. And also Just and Weese Discovering Modern Set Theory (the first volume, anyway). $\endgroup$ – Asaf Karagila Dec 3 '17 at 15:32
  • 1
    $\begingroup$ There is a free preprint copy on his personal webpage. $\endgroup$ – Asaf Karagila Dec 3 '17 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.