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The Cantor set is closed in $[0,1]$ and so its complement in $[0,1]$ should be a countable union of open intervals. Furthermore, every open set containing a point in the Cantor set contains a point not in the Cantor set. Doesn't this give us a correspondence between endpoints of our open intervals and elements of the Cantor set, contradicting countability? Where is this argument going wrong?


marked as duplicate by Asaf Karagila, Andrew D. Hwang, Daryl, Thomas Andrews, vonbrand Mar 2 '14 at 1:41

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  • $\begingroup$ Try Chapter one of Fractal Geometry and Complex Dimensions by Michel Lapidus to understand the complement of the cantor set, namely, the cantor string. $\endgroup$ – Anthony Peter Mar 1 '14 at 23:13
  • $\begingroup$ $1/4$ is in the Cantor set. $\endgroup$ – David Mitra Mar 1 '14 at 23:13
  • $\begingroup$ Can you explain some more what correspondence you are talking about? Each point in the cantor set corresponds to what, exactly? $\endgroup$ – 5xum Mar 1 '14 at 23:14
  • $\begingroup$ Consider the union of the intervals $\left(\frac{1}{2n+1},\frac{1}{2n}\right)$. $0$ is not an endpoint of any of these intervals, but a limit point of endpoints. $\endgroup$ – Daniel Fischer Mar 1 '14 at 23:16
  • $\begingroup$ @Bates : Maybe you didn't notice that the number $1/4$ is a member of the Cantor set, since it's not in any of the excluded middle thirds, but it's also not an endpoint of any of the excluded middle thirds. See my answer below. $\endgroup$ – Michael Hardy Mar 2 '14 at 0:38

Not every point of the Cantor set is an endpoint of an interval of its complement.

But these endpoint of these interval are dense in $C$, and although they are countably many, their closure is uncountable.

There is a good way to convince yourself about this. If you express the elements of $[0,1]$ in the ternary system, i.e., $$ x=\sum_{n=1}^\infty \frac{a_n}{3^n},\,\,a_n\in\{0,1,2\}, $$ then the Cantor set contains exactly those elements in which $a_n\in\{0,2\}$, for all $n\in\mathbb N$. The endpoints of the intervals of the complement contain exactly those point where $a_n=0$ or $2$, but there only finitely many $0$'s or finitely many $2$'s. For example $$ \frac{1}{3}=.02222\ldots,\,\,\,\frac{2}{3}=.20000\ldots,\frac{1}{9}=.00222\ldots, \frac{2}{9}=.02000\ldots,\frac{7}{9}=.20222\ldots,etc. $$ And the elements with only $0$'s and $2$'s in their ternary expansion are as many as the real numbers, while those of them with only finitely many $2$'s are countably many.

  • $\begingroup$ I think the condition should be "all $a_n$ = 0 or 2, with finitely many 0's or finitely many 2's". For example, 1/3 = 0.02222... and 2/3 = 0.20000... (ternary expansions) are both endpoints. $\endgroup$ – Ted Mar 2 '14 at 1:20

No: Most of the points in the Cantor set are not endpoints of deleted middle thirds.

For example, the number $1/4$ is a member of the Cantor set. It's in the lowest third, and in the highest third of that, and in the lowest third of that, and in the highest third of that, and so on, alternating. The number $3/10$ is also in the Cantor set, with a more complicated repeating pattern of upper and lower.

Terminating patterns (all upper or all lower after some number of steps) give you those endpoints;

repeating patterns give you only countably many more points that are not endpoints.

It's the nonrepeating ones that give you uncountably many.


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