The Real Numbers Are Uncountable
Fact: The set of real numbers, ℝ, is uncountable.
Proof: We will focus on the subset (0,1)⊆ℝ and show that this set cannot be countable. Since ℝ sits above an uncountable set it follows that ℝ is also uncountable. We have not yet proven the lemma a set containing an uncountable subset must be uncountable (dear reader, we would also start this by contradiction getting a 1-1 function from the "bigger" set to ℕ).
By contradiction, suppose ∃f:(0,1)↦ℕ which is 1-1. We can now "count" the elements of (0,1): a1=f -1(1), a2=f -1(2),….
We will go further and consider the decimal expansion of each an=0.an1an2an3.... That is, ank is the kth digit of the decimal expansion of an. We put these digits in the the following infinite table where the nth row holds the digits of an.
a11 a12 a13 a14 … a21 a22 a23 a24 … a31 a32 a33 a34 … a41 a42 a43 a44 … ⋮ ⋮ ⋮ ⋮ ⋱
We will now construct a "new" element b∈(0,1) by defining the decimal digits of b=0.b1b2b3b4... as follows. If ann=1, let bn=2. Otherwise, let bn=1. Doing this for every n∈ℕ defines every decimal digit of b and therefore defines b itself. Note also that bn≠ann ∀n.
Now we simply notice that b cannot be in the above "listing" of (0,1) because it cannot be in any of the rows. It cannot be in the jth row because the jth entry of that row is ajj≠bj. In this way we see that b∉{a1,a2,a3,...}=(0,1), which contradicts the construction of b∈(0,1).☐
This fact and its proof can be found in context on my math analysis page.












