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The ordered pair $\{\{a\},\{b,\emptyset \}\}$ seems to be very simple, neat, and highly intuitive ordered pair. So why Kuratowski's pairs were preferred?

bof
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Zuhair
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    My guess is because your proposed definition requires the empty set axiom. It's similar to the alternative definition ${a,{a,b}}$ which requires the regularity axiom. – Alberto Takase Oct 24 '19 at 00:13
  • I prefer ${{a,0},{b,1}}$, but the issue with both of these constructions is that we need to show that $0,1$ exist, which requires the empty set axiom. Kuratowski's construction doesn't require it. Also, I don't see what makes ${b,0}$ simpler than ${b,a}$. – Rushabh Mehta Oct 24 '19 at 00:15
  • @AlbertoTakase, existence of the empty set is a theorem of Zermelo set theory, so it is not an issue here. – Zuhair Oct 24 '19 at 00:21
  • @DonThousand ${b,0}$ has a simpler structure than ${b,a}$ in terms of transitive closures. So why chose the more complex one? – Zuhair Oct 24 '19 at 00:22
  • You already got two answers, and transitive closures aren't a good reason to complicate the essentials of set theory. – Rushabh Mehta Oct 24 '19 at 00:23
  • @Zuhair Correct me if I'm wrong, but did you say the empty set axiom is a theorem? – Alberto Takase Oct 24 '19 at 00:23
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    @AlbertoTakase It can be, depending on the axiomatization. – Rushabh Mehta Oct 24 '19 at 00:24
  • @AlbertoTakase, yes of course. – Zuhair Oct 24 '19 at 00:24
  • @Zuhair It's not obvious at all. What are your axioms you are using? – Rushabh Mehta Oct 24 '19 at 00:24
  • @DonThousand, the usual axioms of Zermelo set theory. They prove the empty set. So stipulating it as an axiom is redundant! – Zuhair Oct 24 '19 at 00:25
  • @AlbertoTakase "In many formulations of first-order predicate logic, the existence of at least one object is always guaranteed. If the axiomatization of set theory is formulated in such a logical system with the axiom schema of separation as axioms, and if the theory makes no distinction between sets and other kinds of objects (which holds for ZF, KP, and similar theories), then the existence of the empty set is a theorem." – Rushabh Mehta Oct 24 '19 at 00:26
  • If you prefer the "existence of a set" over the "empty set axiom" then you are forced to have the separation axioms. But I only know ZFC mainly so my first comment can be disregarded if ZFC is not what's being considered here. – Alberto Takase Oct 24 '19 at 00:27
  • @DonThousand, of course, I'm speaking about the standard line of set theories which are the ones you've mentioned. So there is no problem with the empty set. – Zuhair Oct 24 '19 at 00:27
  • @Zuhair This isn't about whether there is a problem. The main issue is, you can't prove the characteristic property without regularity. That's a bit annoying. – Rushabh Mehta Oct 24 '19 at 00:29
  • @AlbertoTakase, ZFC is an extension of Zermelo. So of course it proves the empty set. ie. its a theorem of ZFC (not an axiom). Existence axioms is part of first order logic so you don't need to add them about sets at all. – Zuhair Oct 24 '19 at 00:30
  • @DonThousand, of course you can! No regularity assumed at all! – Zuhair Oct 24 '19 at 00:30
  • @Zuhair I'd like to see this proof. – Rushabh Mehta Oct 24 '19 at 00:34
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    @DonThousand, that's easy, any two equal pairs must either be both doubletons of a singleton and a doubleton, or doubletons of singletons, or could be singletons. Now with the first situation the singletons cannot be equal to the doubletons (extensionaity) so both singletons must be equal so they must have the same (first projection), same applies to second projection since they are doubletons with a constant in them that is 0. Now for the second situation we'll be having both second projections being 0, so the remainder of them must be equal. the third situation is easy to prove. – Zuhair Oct 24 '19 at 00:49
  • @DonThousand, we don't need to add separation "axioms"; what is needed is to add separation whether axiomatize it or add axioms that prove it, and of course we need the theory to be mono-sorted as you said. For example ZFC formulated as: Extensionality, Regularity, Set Union, Power, Infinity, Choice, and Replacement formalized as $$\forall a,b,c (\phi(a,b) \land \phi(a,c) \to b=c) \to \ \forall A \exists B \forall y (y \in B \leftrightarrow \exists x \in A \phi(x,y))$$ does prove all instances of separation (including the empty set theorem). – Zuhair Oct 24 '19 at 11:21
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    @DonThousand, to me the repetition of $a$ in Kuratowski's pair doesn't make sense, while putting a constant like $0$ would indicate the order of the projection of the pair and it makes full sense. – Zuhair Oct 24 '19 at 11:27
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    As is well known, several def of ordered pair are available in set theory. The popularity of Kuratowski's one is only a "matter of fact" and not a theoretical one. – Mauro ALLEGRANZA Oct 25 '19 at 11:25
  • Maybe the adoption of Kuratowski's def in the 2nd ed (1970) of Bourbaki's Theory of Sets can have had a role in the modern preference (in the 1st edition, Bourbaki introduced ordered pair as a primitive notion). – Mauro ALLEGRANZA Oct 25 '19 at 13:02
  • I wrote a series of blog posts about this that might interest you: 1 2 3 4 – MJD Jan 26 '23 at 15:02

1 Answers1

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This is a nice question, with a perhaps surprisingly subtle answer. Many different set-theoretic implementations of ordered pair have been proposed, especially in the early days of set theory; a good condensed history is given in Kanamori’s paper The empty set, the singleton, and the ordered pair (2003, Bulletin of Symbolic Logic; paywalled there but findable un-paywalled if you search a bit). The first proposal (well, probably equal-first with Hausdorff’s) was Norbert Wiener’s in 1914, defining $(x,y)$ as $\{\{\{x\},\emptyset\},\{\{y\}\}\}$ — similar to your proposal, but with an extra layer of brackets around $x$ and $y$, which I’ll come back to later.

Why did Kuratowski’s emerge as the standard? As discussed in this related question, the choice is to some extent an accident of history — we don’t care about the exact implementation used, so long as it satisfies some essential properties. However, proving those properties will of course depend on what theory/formalism we’re working in — so it’s not just a matter of “anything satisying those properties (in ZF) is equally good”. We have a spectrum of desiderata for any proposed implementation of pairs, from precise mathematical properties through to more context-dependent and subjective criteria:

  1. (essential) for all $x,y$, the specification of $(x,y)$ is a valid definition, i.e. specifies some uniquely existing object;
  2. (essential) for all $x,y,x',y'$, $(x,y) = (x',y')$ if and only if $x = x'$ and $y = y'$;
  3. (near-essential) for any sets $X, Y$, the product set $X \times Y = \{ (x,y) \mid x \in X,\, y \in Y\}$ exists;
  4. (important) the proofs of properties 1–3 should need only a small core fragment of the axioms of set theory, so that it works in other set theories, not just ZFC
  5. (varying with context) the definition should also work in other formalisms, not just ZFC-like first-order theories
  6. (very subjective) general aesthetics: simplicity, symmetry, etc.

Your definition is great on criteria 1–4 — you can prove 1–3 for them with just extensionality, existence of unordered pairs, and the empty set. (And it doesn’t need excluded middle; the proof can be made fully constructive.) And it’s arguably good on criterion 6 too — simpler than Wiener’s, and without the duplication of Kuratowski’s.

However, your proposal falls down on criterion 5. In the modern ZF-style set-theoretic universe, where everything is a set and any of objects can be collected together, it’s fine — but in the early days, that viewpoint wasn’t established immediately, and certainly wasn’t taken for granted until rather later. Much early set theory assumed (first informally, then formalised in systems like Russell’s theories of types) a worldview where there are objects of different types — in the simplest version, levels, with just atoms at the lowest level, then sets of atoms, then sets of sets of atoms, and so on — and a set can only collect together objects from a given type. So forming $\{b,\emptyset\}$ as in your definition is only possible when $b$ is a set — it wouldn’t be admissible if $b$ is an atom, since no sets exist the lowest level, and in particular, no empty set.

This is why Wiener’s version uses $\{\{x\},\emptyset\}$ not just $\{x,\emptyset\}$: he takes singletons to raise all the objects to the same level. That was essential in his original version (A simplification of the logic of relations, 1914, Proc. Cam. Phil. Soc.), since he was working in the typed system of Russell and Whitehead’s Principia. Later, in ZF-style theories, the typing discipline isn’t required, but all other early proposed implementations I’ve seen still follow it.

There have been later definitions which break this typing discipline, such as the definition in Scott and McCarty, Reconsidering ordered pairs (2014, Bulletin of Symbolic Logic); but such definitions are usually proposed when the authors have specific technical requirements that Kuratowski pairing is unsatisfactory (and so are other established implementations) — e.g. in the Scott–McCarty case, to provide pairs of proper classes, not just sets.

So, to summarise:

  • I think early researchers in set theory still preferred to work with notions satisfying the typing discipline of Russell-style systems, even when it was no longer formally required since they’d moved to modern ZF-style systems;
  • your definition doesn’t fit that discipline, which would explain why it wasn’t proposed in the early days;
  • in most later work, the Kuratowski implementation is completely satisfactory and there’s usually no reason to prefer anything else;
  • when people do consider other implementations, it’s for specific technical purposes.
  • Nice! But why it won't work when $b$ is an atom? – Zuhair Jan 25 '23 at 17:38
  • @Zuhair: I’ve edit the answer to explain that a little more! The point is that $\emptyset$ is a set, so it exists at higher levels, but not at the level of atoms. – Peter LeFanu Lumsdaine Jan 25 '23 at 17:45
  • Ah! I see your point now. But, isn't this a little bit artificial. We can simply pick some atom and take $\emptyset$ symbol to denote it instead of denoting the empty set. In reality with such type theories, there is no empty set in the real sense of the word, the empty set is just an arbitrarily chosen empty object and atoms are just arranged to be empty objects distinct from it. – Zuhair Jan 25 '23 at 17:55
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    Thanks for your excellent answer. I would like to add one minor point. Zuhair's proposal, like Kuratowski's, has some degenerate cases which can complicate the proofs of properties 1 and 2. For example, when $a=b=\emptyset$, Zuhair's pair reduces to ${{\emptyset}}$ and one must do a more careful analysis to be sure that property 2 really holds. With Wiener's original definition, the proof is trivial because there are no degenerate cases. The pair of $a=b= \emptyset$ is ${{{a}},{{b},\emptyset}} = {{{\emptyset}},{{\emptyset},\emptyset}}$. – MJD Jan 26 '23 at 14:55
  • But as Wiener said “the particular method selected of doing this is largely a matter of choice”—the specific implementation of pairs is not really that important as long as properties 1–4 can be satisfied. – MJD Jan 26 '23 at 14:57
  • @Zuhair: “In reality with such type theories, there is no empty set in the real sense of the word, the empty set is just an arbitrarily chosen empty object[…]” Not necessarily. Some approaches to theories-with-atoms set things up like that; but in others (e.g. Russell’s), sets and atoms are different types of things entirely, so $\emptyset$ is not arbitrary — it’s the unique empty set. This distinction is essential in constructive set-theories-with-atoms, since you can’t work by cases on whether a set is empty. [cont’d] – Peter LeFanu Lumsdaine Jan 26 '23 at 15:44
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    But that aside — yes, you can tweak your definition to work in typed theories, by replacing $\emptyset$ with some arbitrary object of the same type as $x$ and $y$, in case they’re not sets. That is very similar to Hausdorff’s definition: he used ${{x,0},{y,1}}$, where $0$ and $1$ are arbitrary distinct objects of the type of $x$ and $y$. But then (subjectively) this arbitrary choice slightly detracts from the elegance of the definitions. – Peter LeFanu Lumsdaine Jan 26 '23 at 15:47
  • @MJD: Good point that the singletons in Wiener’s definition avoids degenerate cases. Although in some sense I think the simplification is mostly psychological — looking at fully formal proofs of the injectivity for the various implementations, in the cleanest forms I can find, I don’t think the existence of degenerate cases really adds any extra complications. – Peter LeFanu Lumsdaine Jan 26 '23 at 15:56
  • @PeterLeFanuLumsdaine, MJD, I remember when working on acyclic comprehension in the joint article with Holmes and Bowler this point of degenerate cases was indeed relevant, the Wiener pairs though structurally more complex than Kuratowski, yet the definition of its projections admits a fairly simple characterization. Actually my first proof of equivalence between acyclic and stratified comprehensions used Wiener's pairs. So, these matters can be sometimes relevant. But I agree with Lumsdaine that by the end of the day it doesn't make much difference. – Zuhair Jan 26 '23 at 16:17
  • @PeterLeFanuLumsdaine, I like the answer, but point 5 is a little bit vague to me. I agree that if we take $\emptyset$ to be an arbitrarily chosen atom, then this would render it a version of Hausdorff pairs, and actually this similarity would grow more and more if you extend these pairs to $n$-tuples. You can view this pair at the simplest form of Hausdorff pair (structurally speaking). I'm not that familiar with Russell's type theory. But, if the empty set is stressed as being non-arbitrary set, and there are atoms, then yes this definition of pairs would be at disadvantage. I agree. – Zuhair Jan 26 '23 at 16:31
  • @PeterLeFanuLumsdaine, the tweak for a type theory that starts with a bottom tier of Quine atoms, then the other types are built successively as the sets of the prior type, so all tiers above the bottom one are extensional and have a unique empty set (per type), then this pair would be defined as ${{a^n},{b^n,\emptyset^n}}$ where the superscripts are the types of the superscribed objects. The problem is at type $0$, here $\emptyset^0$ would be some fixed arbitrarily chosen atom, so Hausdorffness only applies to the bottom tier, the higher tiers are not arbitrary. – Zuhair Jan 26 '23 at 16:50