Page 45 - DMTH401_REAL ANALYSIS

Unit 2: Algebraic Structure and Countability

We say that x = y if the digits in their corresponding position in the expansions of x and y are Notes
identical. Thus, if there is even a single decimal places, say, 10th place such that d  c , then
10 10
x  y.

We now discuss the following result due to George Cantor.
Theorem 7: The set of real numbers in the interval ]0, 1[ is not countable.
Proof: Since the set of numbers in ]0, 1[ is an infinite set, therefore, it is enough to how that the set
of real numbers in ]0, 1[ is not denumerable.
If possible, suppose the set of real numbers in ]0, 1[ is denumerable. Then there is a one-one
correspondence between N and the elements of ]0, 1[ i.e. there is a function f: N  ]0, 1[ which is
one-one and onto. Thus, if
f(1) = x , f(2) = x , ........, f(k) = x , ....., then
1 2 k
]0, 1[ = {x , x , ....., x , .....}.
1 2 k
We shall show that there is at least one real number ]0, 1[ which is not an image of any element
of N under f. In other words, there is an element of ]0, 1[ which is not in the list x , x , ....
1 2
Let x , x , ...... be written as
1 2
x = 0, a a a a ....
1 11 12 13 14
x = 0, a a a a ....
2 21 22 23 24
x = 0, a a a a ....
3 31 32 33 34
x = 0, a a a a ....
4 41 42 43 44
...........................................................
...........................................................
...........................................................
From this we construct a real number

z = b b b b .....,
1 2 3 4
where b , b , ..... can take any digits from {0, 1, 2,........., 9} in such a way that b  a , a  a ,
1 2 ] 11 2 22
b = a , ....... Thus,
3 33
z = b b b ....
1 2 3
As a real number in ]0, 1[ such that z  x because b  a , z  x because b  a . In general z  x
1 1 11 2 2 22 n
because a  b . Therefore z is not in the list {x , x x ,....}.
nn n 1 2 3
Hence ]0, 1[ is not countable.
We have already mentioned that the intervals [0, 1], [0, 1[, ]0, 1] and ]0, 1[ are equivalent sets.
Since the set of real numbers in ]0, 1[ is not countable, therefore none of the intervals is a
countable set of real numbers.
Now you can easily conclude that the set of irrational numbers in ]0, 1[ is not countable. If
possible, suppose that the set of irrational numbers in ]0, 1[ is countable. Also you know that the
set of rational numbers in ]0, 1[ is countable and that the union of two countable sets is countable.
Therefore, the union of the set of rational numbers and the set of irrational numbers ]0, 1[ is
countable i.e. the set of all real numbers in ]0, 1[ is countable which by above theorem is not so.
Hence the set of irrational numbers in ]0, 1[ is not countable.
In fact, every interval ]a, b[ or [a, b], ]a, b], [a, b[ is an uncountable set of real numbers.

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