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Topic: How many real numbers are there?
Replies: 18   Last Post: Jul 9, 2006 3:00 PM

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Mike Deeth

Posts: 25
Registered: 12/13/04
How many real numbers are there?
Posted: Jul 7, 2006 4:23 PM
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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 Zermelo-Fraenkel 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

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 well-known 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 :-)

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