The Real numbers

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updates: 29 July 2014

The real numbers

The real numbers is the set = 0 0 where 0 = { alal-1 a1a0 .a-1 a-2 a-3 | l 0, ai {0,1,,9} } 0 = {- alal-1 a1a0 .a-1 a-2 a-3 | l 0, ai {0,1,,9} } and a=b if a-b=0.00000, where addition and multiplication are given by

  1. the mth decimal place of a+b is the same as the mth decimal place of a2m + b2m in ×, and
  2. the mth decimal place of ab is the same as the mth decimal place of a2m b2m in ×.

Define a relation on by xy if y-x0 .

Define the absolute value on |T|: 0 x |x| by |x| = { x, ifx>0, 0, ifx=0, -x, ifx<0.

Define the distance on d:× 0 by d(x,y) = |y-x|.

Let ε>0. The ε-ball at x is ε(x) = { y | d(x,y)<ε} .

Let E be a subset of . The set E is open if

Eis a union of ε-balls. (1.1)

  1. The set with the operations of addition, multiplication, the order and open sets as in (1.1) is an ordered field and a topological field.
  2. The set is a ordered topological subfield of .

Let J with J. The subset J is connected if and only if J is an interval.


: Assume J is not an interval.
Let x,yJ and z with x<z<y,x,yJ andzJ. Let A=(-,z)J and B=(z,)J.
Then A and B are open subsets of J and A,B, AB=andA B=J. So J is not connected.

: Assume J is an interval.
To show: J is connected.
Proof by contradiction.
Assume J is not connected.
Let AJ and BJ be open subsets of J such that AB=, A, Band AB=J. Then f:J{0,1} given by f(z)= { 0, ifzA, 1, ifzB is a continuous surjective function.
Let x1,y1J with f(x1)=0 and f(y1)=1.
Switching A and B if necessary we may assume that x1<y1.
Construct sequences x1,x2, and y1,y2, by xi+1= xi+yi2 andyi+1=yi, iff(xi+yi2)=0, xi+1=xi andyi+1= xi+yi2, iff(xi+yi2)=1. By induction, xiJ and yiJ, and, since J is an interval, xi+yi2J so that f(xi+yi2) is defined and xi+1Jand yi+1J. Also, f(xi+1)=0, f(yi+1)=1, xixi+1<yi+1yi and|xi+1-yi+1| 12|xi-yi| so that |xi+1-yi+1| 12i|x1-y1|. Since is complete and the sequence x1,x2, is increasing and bounded by y1, limnxn exists in .
Since is complete and the sequence y1,y2, is decreasing and bounded by x1, limnyn exists in .
Since limn|xn-yn|=0 then limnxn=limnyn.
Let z=limnxn =limnyn. Since x1x2xn<ynyn-1y1 for n>0 then x1<z<y1. Since J is an interval, zJ.
Since f is continuous, 0=limnf(xn) =f(z)=limnf (yn)=1. This is a contradiction.
So J is connected.

Notes and References

Decimal expansions (real numbers) are introduced to a child to read and write numerical values. Long division (the Eucidean algorithm) is the algorithm that converts rational numbers to real numbers. The integers is the free group on one generator, the rationals is the field of fractions of , and the real numbers is a completion of (a decimal expansion is no different than a Cauchy sequence of rational numbers). Thus, all three number systems are universal objects (in the sense of category theory).

The proof of the theorem that connected sets in are intervals follows the proof given in the course notes of J. Hyam Rubinstein for Metric and Hilbert spaces at the University of Melbourne. This proof does not differ substantially from the proof in [Bou, Gen Top. Ch. IV §2 No. 5 Theorem 4] but is organised to be more self contained.


[BouTop] N. Bourbaki, General Topology, Chapter IV, Springer-Verlag, Berlin 1989. MR?????

page history