Proposition 2.2.1: An Uncountable Set

The open interval (0, 1) is uncountable.
Context

Proof:

Any number x in the interval (0, 1) can be expressed as a unique, never-ending decimal. Actually, this is not quite true: 0.1499999... is the same number as 0.15000.... But when we simply discard those numbers with a non-ending tail of 9's we still get the open interval (0, 1), and now every number has a unique decimal representation. If these numbers were countable, we could list them in a two-way infinite table:

1. number: x₁¹, x₂¹, x₃¹, x₄¹, ...
2. number: x₁², x₂², x₃², x₄², ...
3. number: x₁³, x₂³, x₃³, x₄³, ...
4. number: x₁⁴, x₂⁴, x₃⁴, x₄⁴, ...
...

where each expression in parenthesis represents all decimals in the decimal representation of a particular number without the leading '0.'.

In this list, what would be the number associated to the following element:

Let x be the number represented by (x₁, x₂, x₃, x₄, ...), where we let:
- x₁ = 0 if x₁¹ = 1 and x₁ = 1 if x₁¹ = 0
- x₂ = 0 if x₂² = 1 and x₂ = 1 if x₂² = 0
- x₃ = 0 if x₃³ = 1 and x₃ = 1 if x₃³ = 0
- ...

This new element x is different from the first one in our list, because they differ in their first entry; x is different from the second one in the list, because they differ in the second entry; x is different from the third one because they differ in the third entry, etc. But now it is clear that x can not be in the above list, because it differs with the n-th element of that list in the n-th entry. But this element represents a number in the interval (0, 1). Hence, we have found that we were unable to list all numbers in (0,1), and therefore the interval is indeed uncountable.

Interactive Real Analysis, ver. 1.9.5
(c) 1994-2007, Bert G. Wachsmuth
Page last modified: Mar 28, 2007