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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. – Gerry Myerson Oct 1 '12 at 5:46
@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. – nullUser Oct 2 '12 at 23:09
@null, you may be right. It would be nice of OP to come back to clarify, but perhaps Heisenberg is uncertain. – Gerry Myerson Oct 3 '12 at 1:53

1 Answer 1

I assume that $K_0=[0,1]$, and that for $n\ge 1$, $K_n$ is what’s left after the open middle $(1/3)^n$-intervals have been removed from $K_{n-1}$. If you have already proved that $K_n$ is the union of $2^n$ closed intervals of length $(1/3)^n$, then the first sentence of your argument is fine. The second, however, goes badly astray: you want to show that each point of the Cantor set is the limit of a sequence of endpoints, not that each endpoint is the limit of a sequence of something.

To fix this, let $p$ be any point of the Cantor set. Then for each $n\ge 1$ there is a unique closed interval of length $(1/3)^n$ in $K_n$ that contains $p$. At least one of the endpoints of that interval is an endpoint of a removed interval; choose one that is, and call it $e_n$. (You can’t guarantee that both are: one might be $0$ or $1$.) Now use your first observation to argue that $\langle e_n:n\in\Bbb Z^+\rangle$ is a sequence of endpoints of removed intervals that converges to $p$, so the set of endpoints of removed intervals must be dense in the Cantor set.

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