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Prove that the set of endpoints of removed intervals in the Cantor middle thirds set is a dense subset of the Cantor set.

Attempt at proof:

Since each subinterval is of length $(1/3)^n$, any point contained in $K_n$ is at a distance of less than or equal to $(1/3)^n$ from an endpoint in $K_n$ for all $n$. Thus there exists a sequence of points in $K_n$ that converge to an endpoint in $K_n$.

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    You haven't told us what $K_n$ is, nor what it has to do with the Cantor middle-thirds set.2012-10-01
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    @GerryMyerson I'm pretty sure he means $K_n$ to be the $n$-th approximation of the Cantor set you get, after removing the "middle thirds" $n$ times.2012-10-02
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    @null, you may be right. It would be nice of OP to come back to clarify, but perhaps Heisenberg is uncertain.2012-10-03

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