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The original Zermelo set theory explicitly allowed for urelements.

What was the reason that led Zermelo to formulate the Axiom of Infinity in terms of the existence of a set of the kind that has an infinite rank?

Why didn’t he instead propose the weaker axiom just asserting the existence of a Dedekind-infinite set?

I mean even with modern ZCU minus infinity plus there exists a Dedekind infinite set, call it ZCU-I+I*, this doesn't prove the existence of a set with an infinite rank, we can even add a ranking function to get ZCUR-I+I* + every set has a finite rank, and yet still get to interpret the whole ZCU.

The feeling is that postulating the existence of a set that has infinitely many elements and also has an infinite rank, seems on the face of it to be something stronger than just having a set with infinitely many elements, so what's the benefit from adding infinitute of ranks?

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    $\begingroup$ I'm confused by your question. Zermelo didn't have a notion of rank. He defined a set that is Dedekind-infinite using what he (probably) saw as the simplest construction in set theory. $\endgroup$ Commented yesterday
  • $\begingroup$ @AsafKaragila, I Know, that's why I said "of the kind that", when we look at it from ourdays perspective it is an infinitely ranked set. But, even back then it is obvious that this set has a kind of infinite height in terms of membership steps, so why he thought of such set as a simple kind of infinite set. Since, you are allowing urelements you can easily get a finitely ranked set that is infinite, no need to go up the ladder infinitely many steps go get it. $\endgroup$ Commented yesterday
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    $\begingroup$ But how do you posit "infinite" if your language has no innate function symbols? Is it, perhaps, by considering the function $x\mapsto\{x\}$ and then to ensure that it is not surjective, start with $x=\varnothing$? $\endgroup$ Commented yesterday
  • $\begingroup$ @AsafKaragila, are you saying that at Zermelo's time the set theoretic implementation of ordered pairs was not given yet, his system was posed in 1908, while Weiner and Hausdorff implementations only saw the light six years later. We know nowadays that we can posit "infinite" because all the needed function symbols are defined ones symbolizing known implementation of functions as sets of ordered pairs in the usual sense. So, there is no essential need for the function $x \mapsto \{x\}$. $\endgroup$ Commented yesterday
  • $\begingroup$ I don't see how it would have been beneficial to base set theory intended as foundational on urelements, opaque things about which we can say even less than we can about $\emptyset$. $\endgroup$ Commented yesterday

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The formulation of Zermelo's axiom of infinity i.e., the axiom $\exists s [\varnothing \in s \land \forall x\in s (\{x\} \in s)]$ was informed by Dedekind's informal justification that there are infinitely many "things" in his seminal 1887 essay Was sind und was sollen die Zahlen. In the same essay, Dedekind defines a set $S$ to be infinite provided there is an injection $f:S\rightarrow S$ that is not surjective.

A free copy of an English translation of Dedekind's essay (by Wooster Woodruff Beman) is available here.

The relevant passage in Dedekind's essay are paragraphs 64--66 (on p.31 of the aforementioned English translation).

This MO post is related to the above portion of Dedekind's essay.

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  • $\begingroup$ Yes, but that's different from simply having a set with infinitely many elements? One could have straightforwardly say that there is a Dedekind infinite set without posing a specific set like this. The question is about the infinite rank. $\endgroup$ Commented yesterday
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    $\begingroup$ @ZuhairAl-Johar A specific set has the advantage of giving motivation for a general existential assertion of the type you advocate. Dedekind used this specifc set to motivate his definition of "Dedekind infinite" the same essay; and Zermelo was fully aware of this, given Dedekind's well-established stature and influence, especially on German-speaking mathematicians. $\endgroup$ Commented yesterday
  • $\begingroup$ I see! So that was the reason why. But, looking backwards, it appears to me that matters could have be done by taking the set A of all urelements to be a Tarski infinite set in the sense of not being well order-able by a relation whose converse is a well ordering on it also, then the set of all equivalence classes of finite subsets of A under equivalence relation bijection, could have served as a demonstration of Dedekind's infinity. There is no need to invoke an infinitely ranked set for that purpose. $\endgroup$ Commented yesterday
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    $\begingroup$ @ZuhairAl-Johar Nothing in Zermelo's axioms says anything about either ur-elements or their collection. You're correct that it 'allows for' ur-elements, but it doesn't mandate them. Given that it's explicitly a theory of sets, not of a (potential) set-theoretic universe, it makes sense to say as little about any potential ur-elements as possible. $\endgroup$ Commented yesterday
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    $\begingroup$ @ZuhairAl-Johar perhaps rather than "allows for", it would be better to say "is compatible with". The formal mathematical content of the statement remains the same, while the language makes it sound less like a matter of design which demands attention. $\endgroup$ Commented 18 hours ago

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