prove that the set of real number is denumerable
Answers
he Set of Real Numbers is Uncountable
Theorem 1: The set of numbers in the interval, [0,1], is uncountable. That is, there exists no bijection from ℕ to [0,1].
The argument in the proof below is sometimes called a "Diagonalization Argument", and is used in many instances to prove certain sets are uncountable.
Proof: Suppose that [0,1] is countable. Clearly [0,1] is not a finite set, so we are assuming that [0,1] is countably infinite.
Then there exists a bijection from ℕ to [0,1]. In other words, we can create an infinite list which contains every real number. Write each number in the list in decimal notation. Such a list might look something like:
(1)
1234⋮0.02342424209059039434934...0.32434293429429492439242...0.50000000000000000000000...0.20342304920940294029490...⋮
Let N be the number obtain obtained as follows. For each n∈ℕ, let the nth decimal spot of N be equal to the nth decimal spot of nth number in the list +1 if that number is less than 9, and let it be 0 if that number is equal to 9. In the list above, we would have N=0.1315....
By construction, N is different from every number in our list and so our list is incomplete. But this contradicts the existence of a bijection from ℕ to [0,1]. Hence [0,1] must be uncountable. ◼
Corollary 2: The set of real numbers, ℝ, is uncountable.
Proof: Since [0,1] is uncountable and [0,1]⊂ℝ we have that ℝ is uncountable. ◼
Corollary 3: The set of irrational numbers is uncountable.