The Math Tea argument is an argument that there are real numbers that can’t be defined. The idea is this: there are only countably many definitions of real numbers (e.g., πe or "The middle root of the polynomial x3 − 5x2 + 2x + 4"), and uncountably many real numbers, so there are real numbers that have no definitions.
Elegant as this argument is, it has crucial set-theoretic flaws. For instance, there is no guarantee that there is a set of all the definable real numbers. The axioms of set theory tell us that for any predicate F in the language of set theory there is a set of all the numbers that satisfy F. But the predicate "is definable" is in English, not in set theory.
We can, however, argue for the following weaker claim. Assume set theory is true. Then either:
There is a real number that cannot be defined in the language of set theory, or
"A real number is missing": there is an English language formula F(n) whose only semantic predicate is set-theoretic satisfaction such that there is no real number x whose nth digit after the decimal point is 1 if F(n) and is 0 if not F(n).
Here is the argument. A formula of set-theory defines a real number if it has exactly one free variable and is satisfied by precisely one real number. Say that F(n) if and only if the nth formula of set theory (in lexicographic ordering) defining a real number defines a real number that does not have a 1 in the nth place after the decimal point. The only semantic predicate in F(n) is set-theoretic satisfaction. Suppose (2) is false. Then there is a real number x whose nth digit after the decimal point is 1 if F(n) and is 0 if not F(n). If x can be defined in the language of set theory by a formula ϕ, then suppose ϕ is the nth real-number-defining formula. Then F(n) if and only if x does not have a 1 in the nth place. But x has a 1 in the nth place if and only if F(n). Contradiction! So, x cannot be defined, and hence (1) is true.
Logically speaking, if ZF is consistent, ZFC is consistent both with (1) (this follows by letting the digits of x be defined by the set of all set-theoretic truths and noting that if ZF is consistent, we can consistently suppose there is a set of all set-theoretic truths, but that set of course cannot be defined) and with the denial of (1).
But philosophically speaking, we might reasonably say that (2) would imply that "there aren’t enough real numbers", which sounds wrong, so it seems more reasonable to accept (1) instead.
But if the definable reals don't form a set, then they cannot form R, which is a set. So there must be undefinable reals.
ReplyDeleteHaving a set of all definable reals is just one of the presuppositions of the Math Tea argument. You make a very good point this presupposition is not essential to the argument--I didn't notice that!
ReplyDeleteI should have instead talked of the presupposition that there is a function f from definitions to definable reals such that if D is a definition of x, then f(D)=x. I don't think one can get around that.