Cantor set

Revision as of 21:10, 16 November 2008 by Herefishyfishy1 (talk | contribs) (New page: The '''Cantor set''' is equal to <math>C(0,1)</math>, where <math>C</math> is a recursively defined function: <math>C(a,b)=C\left(a, \frac{2a+b}{3}\right)\cup C\left(\frac{a+...)
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The Cantor set is equal to $C(0,1)$, where $C$ is a recursively defined function: $C(a,b)=C\left(a, \frac{2a+b}{3}\right)\cup C\left(\frac{a+2b}{3},b\right)$ and $C\left(a,a\right)=\{a\}$. Geometrically, one can imagine starting with the set [0,1] and removing the middle third, and removing the middle third of the two remaining segments, and removing the middle third of the four remaining segments, and so on ad infinitum.

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