Today, we're going to have a look at the real numbers from the perspective of the elements of set theory that we have already discussed.

Let us begin with the following theorem due to Cantor.

<THEOREM>

The set of all real numbers is uncountable.

<END THEOREM>

There are many proofs of this theorem.  One notable proof is Cantor's diagonal argument.

This can be found, for example, in Rudin's Principles of Mathematical Analysis.

Here's another:

<PROOF>

Assume that R is countable.

Let <c_n : n < w> be an enumeration of R.

We intend to show that there exists a real number different from each c_n.

We let a_0 = c_0, b_0 = c_k where k is the least k such that a_0 < c_k, a_{n+1} = c_k where k is the least k such that a_n < c_k < b_n, b_{n+1} = c_k where k is the least k such that a_{n+1} < c_k < b_n.

An illustration:

---|---|---|------------|---|---|--- R
  a_0 a_1 a_2  . . .   b_2 b_1 b_0

If we let a = sup {a_n : n \in N}, then a != c_k for every k.

<END PROOF>

The completeness of R is responsible for its uncountability.

We shall not bother to construct the reals from first principles.

Let us however note, for the algebraically inclined, that one constructs the integers from the natural numbers by appropriately defining additive inverses, then takes the field of quotients of the resulting integral domain to construct the rationals, then takes the completion of the rationals to get the reals.

Note also that the set Q of rationals is countable (\aleph_0 . \aleph_0 = \aleph_0).

We have another theorem due to Cantor.

It says that Q is the unique (up to isomorphism) countable dense unbounded linearly ordered set.

A linearly ordered set P is dense if for any a < b there exists c such that a < c < b.

P is unbounded if it has no least and no greatest element.

<THEOREM>

Any two countable linearly ordered sets that are dense and unbounded are isomorphic.

<END THEOREM>

<PROOF>

Let P_1, P_2 be countable dense unbounded linearly ordered sets.

Then we can enumerate P_1 and P_2:

P_1 = {a_n : n \in N}, P_2 = {b_n : n \in N}

It is our task to construct an isomorphism f : P_1 --> P_2.

We do this by defining f(a_0), then f^{-1}(b_0), then f(a_1), then f^{-1}(b_1), etc., in such a way as to keep f order-preserving the whole time.

For example, we can define f(a_n) = b_k where k is the least k such that such that f remains order-preserving.

Hence f(a_0) = b_0, and if say a_1 > a_0, then f(a_1) = b_k where k is the least k such that b_k > b_0, etc.

Note that such a k exists at every step because f is defined thus far for only finitely many elements of P_1, and because P_2 is dense and unbounded.

<END PROOF>

A dense linearly ordered set is complete if every nonempty bounded subset has a supremum.

If P is a dense linearly ordered set, then a set D \subset P is dense in P if for any a < b there is d \in D such that a < d < b.

As we shall now show, the reals are the unique (up to isomorphism) dense unbounded linearly ordered set that is complete and contains the rationals as a dense subset.

<THEOREM> (Completion of dense linear ordering)

Let (P, <) be a dense unbounded linearly ordered set.

Then there is a complete unbounded linearly ordered set (C, <*) such that:

(i) P \subset C, and < and <* agree on P;

(ii) P is dense in C

The set C is unique up to isomorphism; moreover, if C_1 and C_2 are two such sets, then there is an isomorphism h between C_1 and C_2 such that h(p) = p for all p \in P.

<END THEOREM>

<PROOF>

A Dedekind cut in P is a pair (A, B) of disjoint nonempty subsets of P such that:

(a) A \union B = P

(b) a < b for any a \in A and b \in B

(c) If inf B exists, then inf B \in B (note that sup A = inf B).

Let C be the set of all Dedekind cuts in P and let (A_1, B_1) < (A_2, B_2) if A_1 \subset A_2 (and B_2 \subset B_1).

The set C is complete:

If {(A_i, B_i) : i \in I} is a nonempty bounded subset of C, then (\union_{i \in I} A_i, \intersect_{i \in I} B_i) is its supremum.

For p \in P, let A_p = { x \in P : x < p}, B_p = {x \in P : x >= p}

Then P' = {(A_p, B_p) : p \in P} is isomorphic to P and is dense in C.

To prove the uniqueness of C, let C and C' be two complete dense unbounded linearly ordered sets, let P and P' be dense in C and C', resp., and let f be an isomorphism of P onto P'.

Then f can be extended uniquely to an isomorphism f* of C onto C' as follows:

f*(x) = sup {f(p) : p \in P and p <= x}

<END PROOF>

We owe this construction of the real numbers to Dedekind.

Richard Dedekind, "Stetigkeit und irrationale Zahlen", Braunschweig, 1872

The real line, also known as the continuum, is defined as the completion of Q.

Due to this construction, we have that |R| <= 2^{\aleph_0}.

To show that |R| =  2^{\aleph_0}, we have to construct a set of reals of cardinality  2^{\aleph_0}, which we shall now do.

The Cantor Set.

Let C be the set of all reals of the form \sum_{n=1}^{\infty} a_n / 3^n, where each a_n = 0 or 2.

Geometrically, we can form this set by taking the closed interval [0, 1] and removing the middle third, (1/3, 2/3), then removing the middle third from each of the remaining two intervals, and so on.

Then we have that C is in one-to-one correspondence with the set of all w-sequences of 0's and 2's, hence |C| =  2^{\aleph_0}.

The cardinal  2^{\aleph_0} is called the continuum.

The continuum hypothesis, henceforth CH,is the conjecture that  2^{\aleph_0} = \aleph_1.

Neither CH nor its negation is provable in ZFC and we shall return to this idea when we consider forcing.

A real number is called algebraic if it is the root of a polynomial whose coefficients are integers.

A real number which is not algebraic is called transcendental.

Here we have a lemma.

<LEMMA>

If S is a countable set of reals, then |R - S| =  2^{\aleph_0}.

<END LEMMA>

<PROOF>

We rely on a nifty device.

We use R x R instead of R, because |R x R| =  2^{\aleph_0}.

Now, if S \subset R x R is countable, then for some x \in R, S \intersect (R x {x}) = \emptyset, and |R x {x}| =  2^{\aleph_0}.

<END PROOF>

There are then a couple of corollaries:

<COROLLARY>

(a) There are  2^{\aleph_0} irrational numbers.

(b) There are  2^{\aleph_0} transcendental numbers.

<END COROLLARY>

Since Q is dense in R, every nonempty open interval of R contains a rational number.

Hence if S is a disjoint collection of open intervals, S is at most countable.

Let P be a dense linearly ordered set.

P is called separable if P has a countable dense subset.

Hence the reals are separable.

If every disjoint collection of open intervals of P is at most countable, then P is said to satisfy the countable chain condition, henceforth c.c.c.

It follows then that every separable linear ordering satisfies c.c.c.

We are now in a position to state Suslin's problem.

It was first posed by Michail Suslin in Probleme 3, Fund. Math. 1, (1920), 223.

Let P be a complete dense unbounded linearly ordered set that satisfies c.c.c.

Is P isomorphic to the reals?

It turns out that this is not decidable in ZFC.

One day, in the distant future, we shall return to this question.

I should like to have a guest lecturer in the form of an active researcher in set theory to explain it to us.

One can only hope.