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Sat, 13 Nov 2021
I said it was obvious but it was false
In a recent article about Norbert Wiener's construction of ordered pairs, I wrote:
I said “it's obvious that both are correct”, but actually no! The first one certainly seems correct, but when I thought about it more carefully, I realized that the first formula is extremely complex, vastly moreso than the second. That !!|x|!! notation, so intuitive and simple-seeming, is hiding a huge amount of machinery. It denotes the cardinality of the set !!x!!, and cardinality is not part of the base language of set theory. Normally we don't worry about this much, but in this context, where we are trying to operate at the very bottom bare-metal level of set theory, it is important. To define !!|x|!! requires a lot of theoretical apparatus. First we have to define the class of ordinals. Then we can show that every well-ordered set is order-isomorphic to some ordinal. Zermelo's theorem says that !!x!! has a well-ordering !!W!!, and then !!\langle x, W\rangle!! is order-isomorphic to some ordinal. There might be multiple such ordinals, but Zermelo's theorem guarantees there is at least one, and we can define !!|x|!! to be the smallest such ordinal. Whew! Are we done yet? We are not. There is still more apparatus hiding in the phrases “well-ordering” and “order-isomorphic”: a well-ordering is certain kind of relation, and an order isomorphism is a certain kind of function. But we haven't defined relations and functions yet, and we can't, because we still don't have ordered pairs! And what is that !!\langle x, W\rangle!! notation supposed to mean anyway? Oh, it's just the… um… whoops. So the !!|x|=1!! idea is not just inconveniently complex, but circular. We can get rid of some of the complexity if we restrict the notion to finite sets (that eliminates the need for the full class of ordinals, for the axiom of choice and the full generality of Zermelo's theorem, and we can simplify it even more if we strip down our definition to the minimum requirements, which is that we only need to recognize cardinalities as large as 2. But if we go that way we may as well dispense with a general notion of cardinality entirely, which is what the second definition does. Even without cardinality, we can still define a predicate which is true if and only if !!x!! is a singleton. That's simply: $$\operatorname{singleton}(x) \equiv_{def} \exists y. x = \{y\}.$$ One might want to restrict the domain of !!y!! here, depending on context.¹ To say that !!x!! has exactly two elements² we proceed analogously: $$\exists y. \exists z. z\ne y \land x = \{y, z\}.$$ And similarly we can define a predicate “has exactly !!n!! elements” for any fixed number !!n!!, even for infinite !!n!!. But a general definition of cardinality at this very early stage is far out of reach. [ Addendum 20220414: I have no idea what I was thinking of when I said “… even for infinite !!n!!” in the previous paragraph, and I suspect that it is false. ] ¹ This issue is not as complex as it might appear. Although it appears that !!y!! must range over all possible sets, we can sidestep this by restricting our investigation to some particular universe of discourse, which can be any set of sets that we like. The restricted version of !!x\mapsto |x|!! is a true function, an actual set in the universe, rather than a class function. ² There does not seem to be any standard jargon for a set with exactly two elements . As far as I can tell, “doubleton” is not used. “Unordered pair” is not quite right, because the unordered pair !!\{a, a\}!! has only one element, not two. [Other articles in category /oops] permanent link |