9. continued- The Continuum of Real Numbers

The Real Numbers

Since we are particularly interested in the cardinality of the Reals, it turns out to be notationally simpler to work with the "open" unit interval 0 $\QTR{Large}{<r<\ }$ 1, , written $\QTR{Large}{(}$ 0,1 $\QTR{Large}{)}$ .

9.5 Lemma

0,1 $\QTR{Large}{))}$

Proof:

This amounts to producing the appropriate 1 to 1 onto map of Dedekind cuts. 0,1 $\QTR{Large}{)}$ . For clarity, this can be done as a composition, $\QTR{Large}{f=gh}$

-1,1 $\QTR{Large}{)}$ where $\QTR{Large}{h(}$ x $\QTR{Large}{)=}$
$\QTR{Large}{g:(}$ -1,10,1 $\QTR{Large}{)}$ where $\QTR{Large}{g(}$ x $\QTR{Large}{)=}$

We will want to calculate the cardinality of the Continuum of Real Numbers. We begin by identifying the set of Dedekind cuts in $\QTR{Large}{(}$ 0,1 $\QTR{Large}{)}$ with a particular set of infinite decimal fractions. Explicitly,

Let $\QTR{Large}{D=\{}$ 0,1,2,3,4,5,6,7,8,9 $\QTR{Large}{\}.}$ Note that applying 9.4, gives

Let be all maps not excluded by the following two bullets.

The constant map $\QTR{Large}{f(}$ x $\QTR{Large}{)=\ }$ 9 is not in $\QTR{Large}{F.}$
$\QTR{Large}{f(}$ x $\QTR{Large}{)}$ is not in $\QTR{Large}{F}$ if there exists an n such that $\QTR{Large}{f(}$ x $\QTR{Large}{)=0}$ for all x $\QTR{Large}{>\ }$ n.

Notationally, given we write $\QTR{Large}{f(}$ i $\QTR{Large}{)=}$ d We are identifying $\QTR{Large}{f}$ with the infinite

decimal fractions, .ddd......d.....

Where not all the d9 and the sequence does not end in an infinite string of 0's.

Definition

We define a map (0,1) as follows:

Given ,let d(i)=.ddd......d , the ith. (finite) partial fraction.

Let $\ ,$ where, d(i) for some i $\QTR{Large}{\}.}$

9.6 Lemma:

(0,1) is one to one and onto, hence (0,1) $\QTR{Large}{).}$

Proof:

As before, but note that in the definition of $\QTR{Large}{F}$ , had we not excluded .99999999..., $\QTR{Large}{q}$ would have produced $\QTR{Large}{.}$

The Cardinality of the Continuum of Real Numbers:

In the text, the notation "c" is used for . In these notes we will stick to the second notation.

The proof of the following theorem turns out to be deceptively straight forward, given the material that we will discuss on Page 14.

9.7 Theorem:

Proof:

Using 9.5 and 9.6, it suffices to show, , which in turn ,applying 9.1, requires producing 1 to 1 maps and .

One direction is trivial since inclusion gives a one to one into map

Next let $\QTR{Large}{E=\{}$ 1,2,3,4,5,6,7,8 $\QTR{Large}{\}}$

Apply 9.5 to conclude that , hence there is a 1 to 1 onto map Finally, composing this with the inclusion gives a 1 to 1 map

The Continuum Hypothesis:

From the point of view of set maps we have natural strict inclusions. We have to make a choice for We have shown

It is not difficult to find other sets between and of size The natural question to ask is are other cardinalities between and ? In particular, can we find a set $\QTR{Large}{S}$ such that

and

Cantor conjectured that this was not possible. We shall see that this conjecture has has resulted in some very interesting mathematics.

Evidence - Measure

Working Definition - Let We say that has measure 0, if for any there exists a, possibly countable, set of closed intervals of length greater than 0 such that

MATH and

1 . If then $\QTR{Large}{S}$ has measure 0.

choose some 1 to 1, onto map . For each $\QTR{Large}{i}$ choose such that and

note that MATH and

For example if , and is the irational in some ordering of the rationals by then set

2. The Cantor Middle Third Set -

Take the closed unit interval [0,1] and delete the open middle third (1,3)

Call the resulting set $\QTR{Large}{C}_{1}$ . note that the "length" of $\QTR{Large}{C}_{1}$ , equal the sum of the lengths of the intervals in $\QTR{Large}{C}_{1}$ , is

Suppose we have defined $\QTR{Large}{C}_{n}$ as a union of closed intervals in [0,1], and that the length of $\QTR{Large}{C}_{n}$ is .

Define to be the result of deleting the open middle third out of each of the intervals in $\QTR{Large}{C}_{n}$ .

Note that the length of is

Finally, define MATH Almost by definition, $\QTR{Large}{C}$ has measure 0. On the other hand

Check that $\QTR{Large}{C}$ is more or less, 0,2 $\QTR{Large}{\})}$ ,using base three arithmetic.