# Cantor Set proofs

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 $S$.

• Self work, ideas, insights...??? – DonAntonio Oct 28 '12 at 4:32
• When I wrote the above I meant actual self work, not to write down the definition of the Cantor set... – DonAntonio Oct 28 '12 at 4:43
• What properties of compact sets do you know? – user642796 Oct 28 '12 at 4:46
• In your idea for solving (c) I expect you mean $1 / 3^N$ and not $1 / 3N$. – user642796 Oct 28 '12 at 5:16

a) There are two approaches. First you can explicitly show that for example $\frac{1}{3}$ is an element of every $S_k$ and hence is contained in the Cantor set.

If you want a more theoretical argument, then we would apply the Cantor intersection theorem as you said. Note that $$S_0 \supseteq S_1 \supseteq S_2 \supseteq \cdots$$ Each $S_k$ is a compact set (why?) and use these facts to apply the intersection theorem.

b) Note that $C \subseteq S_0$ which is bounded. You also need to prove that the set is closed.

c) $\mathbb{R}$ is a connected space and so the only sets which are clopen are $\mathbb{R}$ itself and $\emptyset$. From a we know the set is non-empty. From b we know the set is compact (and hence closed). What can you conclude?

a) You can check that $0$ is in each $S_n$, which means $0 \in S$, so $S$ is nonempty.

b) Since $S_1$ is closed, then by induction we can show that each $S_n$ is closed. Also, we know it is bounded since it's in $[0,1]$. Hence, you have it is compact in $\mathbb{R}$.

c) Check that $1/3$ is in $S$, but every open interval around $1/3$ contains something in the interval $(1/3, 2/3)$ which is outside of $S$. Hence, $1/3$ is in $S$ but is not an interior point of $S$, so $S$ is not open. Indeed, $S$ has empty interior.