
How many real numbers are there?
Posted:
Jul 7, 2006 4:23 PM


Answer: a countable infinitude.
Proof: Well, before we start, the reader will recall a construction of the real numbers. The two most common constructions are dedekind cuts and appropriate equivalence classes of Cauchy sequences of rationals. The important point is not the specific construction, but the fact that a given real number is nothing more than a single ZermeloFraenkel set. For example, if we use the Dedekind cut construction, then the real number pi is literally the ZF set of rationals which are at most pi (where "rational" and "at most" are both in turn defined using ZF or ZFC).
Let X_1 be the set of sets which can be constructed from the ZF axioms in at most 1 step and using only logical sentences of length 1 or less. In general, let X_n be the set of sets which can be constructed from the ZF axioms in at most n steps using logical sentences of length n or less. Alternatively (and this is being pedantic, but I must be pedantic since my detractors will lunge at any slightest gap in my reasoning), we can say X_1 = { emptyset }, and, having defined X_i, we define X_{i+1} by saying S in X_{i+1} iff one or more of the following hold:
1. (separation) S = { x in A  P(x) } where A in X_i and P is a logical sentence in x of length at most i+1 2. (union) S = A union B where A,B in X_i 3. (pairing) S = {A,B} where A,B in X_i 4. (replacement) S = {y in B  exists x in A s.t. P(x,y)} where A,B in X_i and P is a logical sentence in x,y of length at most i+1 5. (infinity) S = { emptyset, {emptyset}, {{emptyset}}, ... } 6. (power set) S = power set of A, where A in X_i
One could also add a "choice" condition, but we'll leave it out since it plays no role and the argument would be identical anyway.
It's easy to get a bound on S_i. We won't explicitly calculate one, because it depends on the first order language being used in the logic sentences, but obviously each S_i < infty.
It's wellknown that a countable union of finite sets is countable. So the set T = S_1 union S_2 union ... is countable. Let r be a real number. By our above observations, r is just some ZF set, and hence r in S_i for some i, hence S in T.
>From here I'll assume the reader believes that subsets of a countable set are countable. This should be proved, but I am running short on time... I'm sure my critics will clap their hands at this and heave a sigh of relief as it gives them a straw man to let them continue their Cantorish illusions just a little longer, but I guess it doesn't matter much since even if I filled in every minute detail, they'd just hurl personal insults and so forth then.
Anyway, I hope this letter finds you in good health! Nathan The Great, Age 11 :)

