Cantor SetYou are currently
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It is formed by taking the closed interval [ 0, 1 ] and repeatedly taking the middle third from any line segment.
Therefore at the first iteration the open set ( 1/3, 2/3) is removed.
At this point the set consist of the union of four closed intervals namely [0, 1/9 ] U [ 2/9, 1/3 ] U [ 2/3, 7/9 ] U [ 8/9, 1 ]
Consider all numbers in the interval [ 0, 1] written in tertiary (base 3).
At the nth iteration only numbers with 1 in the nth tertiary place will be removed.
Show that 1/4 is not an end point of intervals.
Counting up the length of the interval that are removed at each iteration.
1/3 + 2/9 + 4/27 + 8/81 .... = 1
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