Prove. The closed interval U = [0, 1] is non-denumerable.
Proof. Let us assume the contrary. Assume the interval U = [0, 1] is a denumerable set. Then all of its points can be arranged in a sequence
1) x1, x2, x3, ......
We thus assume that every point x ε U occurs in the sequence 1). Now divide U into three equal parts by means of the points 1/3 and 2/3 giving the intervals
Now x1 cannot belong to all three intervals (if it is an endpoint it could belong to two of the intervals). Denote by U1 the interval that does not contain x1. Now divide the interval U1 into three sub-intervals of equal length and denote by U2 the interval that does not contain x2. Now divide U2 into three sub-intervals of equal length and denote by U3 the interval that does not contain x3. If we continue in this manner we obtain an infinite sequence of nested intervals
which possesses the property that
We thus end up with an interval Un that contains no points. Now the length of the interval Un is 1/3n so it is clear that this length approaches zero as n . Now this result that we have obtained, an interval Un that contains no points, stands in contradiction to the result predicted by Cantor’s Principle which states that there will be one point ζ belonging to all the intervals in such a nested system. We have thus reached a contradiction and conclude that our hypothesis that the interval U = [0, 1] is a denumerable set is false. Thus it must be non-denumerable.