The Cantor Set is constructed in the following way:
Start with the interval [0,1]. Next, remove the open middle third interval, which gives you two line segments [0,1/3] and [2/3,1]. Again, remove the middle third for each remaining interval, which leaves you now with 4 intervals. Repeat this final step ad infinitum.
The points in [0,1] that do not eventually get removed in the procedure form the Cantor set.
How many points are there in the Cantor Set?
Consider the diagram below:
An interval from each step has been coloured in red, and each red interval (apart from the top one) lies underneath another red interval. This nested sequence shrinks down to a point, which is contained in every one of the red intervals, and hence is a member of the Cantor set. In fact, each point in the Cantor set corresponds to a unique infinite sequence of nested intervals.
To label a point in the Cantor set according to the path of red intervals that is taken to reach it, label each point by an infinite sequence consisting of 0s and 1s.
A 0 in the nth position symbolises that the point lies in the left hand interval after the nth stage in the Cantor process.
A 1 in the nth position symbolises that the point lies in the right hand interval after the nth stage in the Cantor process.
For example, the point 0 in [0,1] is represented by the sequence 0000…., the point 1 is represented by the sequence 1111…. and the point 1/3 is represented by the sequence 01111….
So, as there are infinite sequences consisting of 0s and 1s, there are an infinite number of elements in the Cantor set. If we place a point before any one of these infinite sequences, for example 0100010… becomes .0100010…, then we convert an infinite sequence of 0s and 1s to the binary expansion of a real number between 0 and 1. This means that the number of points in the Cantor set is the same as the number of points in the interval [0,1]. We conclude that the infinite process of removing middle thirds from the interval [0,1] has no effect on the number of points in [0,1]!