Let $S_0=[0,1]$ and let $S_k$ be defined in the following manner for $k\geq 1$: \begin{align*} S_1&=S_0-\left(\frac{1}{3},\frac{2}{3}\right)=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\ S_2&=S_1-\left\{\left(\frac{1}{9}, \frac{2}{9}\right)\cup \left(\frac{7}{9}, \frac{8}{9}\right)\right\}=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\ S_3&=S_2-\left\{ \left(\frac{1}{27}, \frac{2}{27}\right)\cup \left(\frac{7}{27}, \frac{8}{27}\right)\cup \left(\frac{19}{27}, \frac{20}{27}\right)\cup \left(\frac{25}{27}, \frac{26}{27}\right) \right\}\\ &=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27}, 1\right]\\ \vdots \end{align*} Then put $C=\bigcap_{k=0}^\infty S_k$. This set is known as the Cantor set. a.) Prove that $C$ is nonempty. b.) Prove that $C$ is compact. c.) Prove that $C$ is not an open set.
My knowledge of a.) is that Cantor's intersection theorem is closely related to the Heine-Borel theorem and Bolzano-Weierstrass theorem, each of which can be easily derived from either of the other two. Hence, some how we these can be used to show that the Cantor set is nonempty. However, I can't figure out how to set this proof up.
For b.) I need to show that $S$ is totally bounded, then I could use Heine–Borel theorem to say it is compact. Once again, I need a little help setting it up.
Does this logic work for c.)? Now suppose that there is an open set $U$ contained in $S$. Then there must be an open interval $(a, b)$ contained in $S$. Now pick an integer $N$ such that $1 / 3 N < b - a$. Then the interval $(a, b)$ can not be contained in the set $AN$, because that set is comprised of intervals of length $1 / 3N$. But if that interval is not contained in $AN$ it can not be contained in $S$. Hence, no open set can be contained in the Cantor set $S$.