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Describe the interior of Cantor set

I think the interior is empty because the cantor set of nowhere dense, but as I write it correctly?

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well the cantor set contains no interval, so if the interior were not empty then.... –  user38268 Apr 18 '12 at 14:01
    
Since the Cantor set is a set of points, wouldn't "What is the interior of a point?" be an equivalent question? –  draks ... Apr 18 '12 at 14:22
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@draks: W... What? –  André Caldas Apr 18 '12 at 14:26
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@draks: $[0,1]$ is a set of points too, and that set of points certainly has an interior. –  Henning Makholm Apr 18 '12 at 14:41
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@draks: Both are sets of points. You may be looking for a "set of isolated points" -- but the Cantor set is not such a set. –  Henning Makholm Apr 18 '12 at 14:54

1 Answer 1

up vote 7 down vote accepted

The Cantor set is the intersection of the sets $F_i$, where

$F_1=[0,1/3] \cup [2/3,1]$,

$F_2=[0,1/9]\cup [2/9,1/3]\cup [2/3, 7/9]\cup [8/9,1]$,

and, in general: if $F_{n}$ is defined and consists of the union of $2^n$ disjoint intervals of the form $[k/3^n, (k+1)/3^n]$, then $F_{n+1}$ is obtained by removing the "open middle third" of each of these intervals.

Note the total length of $F_n$ is $(2/3)^n$.

Suppose the non-empty open interval $I=(a,b)$ is contained in the Cantor set. Then for each $n$, $I$ is contained in $F_n$, and thus in one of the intervals defining $F_n$. But then $0< b-a\le (2/3)^n$ for all $n$. This leads to a contradiction, as $b-a\gt0$ and $\lim\limits_{n\rightarrow\infty} (2/3)^n=0$.

So the Cantor set contains no non-empty open interval; thus the interior of the Cantor set is empty.

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