Related to a question I asked earlier.:
Let $F$ be the subset of $[0,1]$ constructed in the same manner as the Cantor set except that each of the intervals removed at the $n$th iteration has length $\frac{\alpha}{3^{n}}$ with $0 < \alpha < 1$.
I need to show that $F$'s complement, $[0,1]\backslash F$ is dense in $[0,1]$.
I know that a $[0,1]\backslash F$ is dense in $[0,1]$ if any open interval in $[0,1]$ contains a point of $[0,1]\backslash F$.
Obviously, the open intervals in $[0,1]$ comprising $[0,1]\backslash F$ contain points $[0,1]\backslash F$,so, I just need to worry about the open intervals inside $[0,1]$ that are not part of $[0,1]\backslash F$.
However, those would be open intervals inside the Cantor set, and while I know that the Cantor set does not contain any intervals of positive measure, I don't know how to show this all epsilon-y (i.e., with neighborhoods, ets). Therefore, I was asking how to formally prove this (either an outline or a full proof is fine; hints are fine, too, if you don't mind follow-up questions.)
Best Answer
Let $F_i$ be the result of the $i$-th iteration of removing all the $\alpha/3^i$ wide open intervals. That means $F_0 = [0,1]$, $F_1 = [0,\frac12-\frac\alpha6]\cup [\frac12 + \frac\alpha6, 1]$ and so on. We then have $F = \bigcap_i F_i$.
Now, every $F_i$ consists of $2^i$ disjoint, closed intervals of equal length. That means that each such closed interval is less than $1/2^i$ wide.
Assume $(a, b) \subseteq F$ with $a < b$. That means that $(a, b) \subseteq F_i$ for all $i$, by definition of $\bigcap$. But that can't be true, since $b-a > 1/2^k$ for some $k$, and in the corresponding $F_k$ there is no room for an interval of that width.
Therefore there can be no non-empty open interval in $F$, which makes the complement of $F$ dense.