How exactly do the ZFC axioms guarantee the existence of ordered pairs or exactly which axioms guarantee this and how? I've googled this question for 15 minutes but haven't found anything that remotely answers this. Thanks in advance.
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8Wow... 15 whole minutes? Incredible! – Asaf Karagila Jun 26 '17 at 21:25
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2https://math.stackexchange.com/questions/141646/proving-the-pairing-axiom-from-the-rest-of-zf and https://math.stackexchange.com/questions/62908/how-can-an-ordered-pair-be-expressed-as-a-set/ combined. – Asaf Karagila Jun 26 '17 at 21:27
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It's a straightforward application of the pairing axiom.
Suppose I have two sets $a, b$. Then:
$(i)$ The set $x=\{a, a\}=\{a\}$ exists by pairing. Note: the pairing axiom says $\forall x\forall y\exists z\forall w(w\in z\iff (w=x\vee w=y))$, and this nowhere requires $x\not=y$, so forming the pair $\{a, a\}$ is perfectly legal.
$(ii)$ The set $y=\{a, b\}$ exists by pairing.
$(iii)$ So the set $\{x, y\}$ exists by pairing. But this is just $\{\{a\}, \{a, b\}\}$, that is, $\langle a, b\rangle$.
This uses the Kuratowski definition of ordered pairs. There are other ways you can define ordered pairs (the ordered pair isn't a primitive concept in set theory), and if you want to use a different definition then the proof will be somewhat different. But for all the reasonable definitions of ordered pairs, it looks pretty much exactly the same as the above.
EDIT: As Asaf points out, the pairing axiom can easily be proved from the other ZFC axioms. I find it valuable to explicitly include it anyways, since I (and others) are frequently interested in subtheories of ZFC which contain the pairing axiom but don't contain the other axioms used to prove it.
Noah Schweber
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Thank you so so much for your clear 3-step explanation! Now it's very clear. I was suspecting the existence of ordered pairs was related to the Axiom of Pairing due to the Wikipedia entry on the Axiom of Pairing, but I didn't fully understand its explanation. – user115859 Jun 26 '17 at 21:38
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This results from the axiom of pairing: an ordered pair is defined as $$(a,b)\stackrel{\text{def}}{=}\bigl\{\{a\},\{a,b\}\bigr\}.$$
Bernard
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The axiom of pairing is not entirely standard and many axiomatizations omit it, as it follows from power set and replacement. – Asaf Karagila Jun 26 '17 at 21:27
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@AsafKaragila Really? I've never seen pairing omitted from the ZFC axioms. (But I'm biased - I'm very interested in systems with little or no replacement, and even without powerset, so to me using those to get pairing feels very wrong.) – Noah Schweber Jun 26 '17 at 21:29
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@Noah: I've never seen it not omitted, actually. Even if it was introduced at first, it was always omitted within one hour as an exercise in working with the axioms. – Asaf Karagila Jun 26 '17 at 21:29
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@AsafKaragila Wikipedia lists it, as do (more seriously) Kunen (page 12) and Jech. – Noah Schweber Jun 26 '17 at 21:31
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@Noah: In the lecture notes I wrote for the course I gave, Pairing was an exercise that follows from the axioms (I recall doing it in class). There was also an exercise explaining that Empty Set and Separation are redundant in presence of Infinity and Replacement. The thing is that when you aim to check that each axiom holds in some model, it's easier to have less axioms (usually), especially when some axioms almost trivially follow from the others. I agree that some redundancy is good, like Separation (which later can be used for bounded separation and so on), but pairing? That's too much... – Asaf Karagila Jun 26 '17 at 21:38
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@AsafKaragila I definitely like having the exercise of proving it from the others, but I also like having it on its own at the beginning - it lets us prove basic facts without leaping into the "content-ful" axioms (powerset, separation, replacement), which I think is helpful for learning the system. (Also see my previous comment re: interest in subsystems.) – Noah Schweber Jun 26 '17 at 21:40
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@Noah: I definitely agree, and I think this is a matter of "interest" here. If you're teaching ZFC in order to focus on its subsystems, then more axioms might be better. If you're teaching ZFC as a first course in axiomatic set theory, and want to explore that specific system, then more axioms isn't necessarily a good thing. – Asaf Karagila Jun 26 '17 at 21:41