REAL NUMBER SET IS UNCOUNTABLE
HAO LIU
We prove the real number set is uncountable mainly by making use
of the so-called Cantor Diagonalization Argument. Specifically we construct a real number in (0, 1) that cannot be represented by a supposed
bijective function f from N to (0, 1) by this argument. Genious this
method looks, we still have a couple of alternative approaches to show
the real number set is uncountable. The proof below is one of the few.
Note: We consider it trivial to prove the real number set is not finite.
Hence mainly we’ll concentrate on the non-denumerable proof.
Claim: The set of all real numbers is uncountable.
Proof. : We’ll show that R is not denumerable by contradiction.
Suppose that R is denumerable, i.e., there exists a bijective funtion
f : N → R.
Then consider the following two sequences defined in a recursive manner:
a1 = f (1)
b1 = f (min{i ∈ N | f (i) > a1 })
an+1 = f (min{i ∈ N | an < f (i) < bn })
bn+1 = f (min{i ∈ N | an+1 < f (i) < bn })
Notice that
(1)
∀n ∈ N, an < an+1 < bn+1 < bn .
Next we’ll show that ∀n, m ∈ N, an < bm .
Suppose ∃n, m ∈ N such that an ≥ bm . Then since ∀n ∈ N, an < bn ,
we have only two cases: n < m or n > m.
(i) Case 1: if n < m then by (1) we know am > an ≥ bm . A contradiction with am < bm .
(ii) Case 2: if n > m then by (1) we obtain that an ≥ bm > bn . A
contradiction with an < bn .
Date: November 25, 2009.
Key words and phrases. Real Number Set, Uncountable.
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HAO LIU
Therefore
∀n, m ∈ N, an < bm .
(2)
Let c = sup{an | n ∈ N}. Since {an } is nonempty and bounded, it
has a supremum, and we denote it as c. Next we’ll show ∀n ∈ N, c < bn .
Suppose ∃n ∈ N such that c ≥ bn . Then by (1), we have bn+1 < bn ≤ c.
Then by the definition of supremum, ∃m ∈ N such that bn+1 < am < c.
This is a contradiction with (2).
Therefore we have
(3)
∀n, m ∈ N, an < c < bm .
Notice that c is a real number and since f is surjective, we know that
∃k ∈ N such that f (k) = c. Next we’ll show there exists a contradiction.
By the definition of an , we obtain ∀n ∈ N, ∃ln ∈ N such that an =
f (ln ). Then by the definition of an , we have
an = f (ln ) = f (min{i ∈ N | an−1 < f (i) < bn−1 }).
Since f is injective we have ln = min{i ∈ N | an−1 < f (i) < bn−1 }.
Also notice that an−1 < f (k) = c < bn−1 . Hence ln ≤ k.
By the similar argument, we note that an−1 < an+1 = f (ln+1 ) < bn−1 .
Then it follows that ln ≤ ln+1 . But since an = f (ln ) < an+1 = f (ln+1 )
and f is injective, therefore ln 6= ln+1 .
Thus we have
(4)
ln < ln+1 ≤ k, ∀n ∈ N
But since k < ∞, l1 ≥ 1 and ln+1 ≥ ln + 1, ∀n ∈ N, it follows that
∀n > k, ln ≥ ln−1 + 1
≥ ln−2 + 2
≥ · · · ≥ l1 + n − 1
≥n>k
A contradiction with (4).
Therefore we have reached an unavoidable contradiction. Hence R is
non-denumerable.