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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) x_{1}, x_{2}, x_{3}, ......

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 x_{1} cannot belong to all three intervals (if it is an endpoint it could belong to two of the
intervals). Denote by U_{1} the interval that does not contain x_{1}. Now divide the interval U_{1} into
three sub-intervals of equal length and denote by U_{2} the interval that does not contain x_{2}. Now
divide U_{2} into three sub-intervals of equal length and denote by U_{3} the interval that does not
contain x_{3.} 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 U_{n} that contains no points. Now the length of the interval U_{n} is
1/3^{n} so it is clear that this length approaches zero as n
. Now this result that we have
obtained, an interval U_{n} 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.

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