Wikipedia 10K Redux by Reagle from Starling archive. Bugs abound!!!

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# CantorSet

```The Cantor Set is a remarkable construction involving only the real numbers between zero and one.  The concept is simple, you take the interval (0,1) and remove the "middle third" of this interval.  Then you take what is left (0, 1/3) U  (2/3, 1) and remove the "middle thirds" from these two intervals.  Continue this process ad infinitum.  Pictorially, the process would look something like this:

0                 1/3                  2/3                 1
============================================================
====================                    ====================
=======      =======                    =======      =======
==   ==      ==   ==                    ==   ==      ==   ==

The question becomes, what is left when you are done?  If you add up the length of segments removed, it would calculate out to be:

1/3 + 2/9 + 4/27 + 8/81 + ... + 2^n/3^(n+1) = (1/3) SUM [n=1,infinity] (2^n/3^n)
= (1/3) (1 / (1- 2/3))
= (1/3) (1 / (1/3))
= (1/3) (3)
= 1

Using this calculation, you may be surprised if there were anything left - after all, the sum of the lengths of the removed intervals is equal to the length of the original interval.  However a closer look at the process reveals that we must have something left, since removing the "middle-thirds" of an interval involved removing open sets (sets that do not include their endpoints).  So removing the line segment (1/3, 2/3) from the original (0,1) interval leaves behind the points 1/3 and 2/3.  A little reflection will convince you quickly that they will never be removed, in fact none of the enpoints of any of the intervals at any stage in the process will ever be removed.  So we know for certainty that the Cantor set is not empty.

In fact, it can be shown that there as many points left behind in this process as there were those that were removed.  To see this, consider the points in the (0,1) interval in terms of base-three notation.  In this notation, 1/3 can be written as 0.1 and 2/3 can be written as 0.2.  If we remove everything from 1/3 and 2/3 we are really removing everything between 0.1 and 0.2, or in other words, '''everything with a 1 in the first position after the point''' (except for .1 itself).  The next step examines the intervals (0, 0.1] and [0.2, 1) and removes their middle thirds.  In this case we are removing everything between 0.01 and 0.02 in the first interval and between 0.21 and 0.22 in the second interval.
```