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