real analysis - Induction proof showing the Cantor set is closed.

Definition: A set $S \in \mathbb{R}$ is open if its complement is closed.

Let the Cantor set be defined as $C=\bigcap_{k=0}^\infty S_k$ and let $S_0=[0,1]$ and let $S_k$ be defined in the following manner for $k\geq 1$.

$$\begin{align*} S_1&=S_0-\left(\frac{1}{3},\frac{2}{3}\right)=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\ S_2&=S_1-\left\{\left(\frac{1}{9}, \frac{2}{9}\right)\cup \left(\frac{7}{9}, \frac{8}{9}\right)\right\}=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\ S_3&=S_2-\left\{ \left(\frac{1}{27}, \frac{2}{27}\right)\cup \left(\frac{7}{27}, \frac{8}{27}\right)\cup \left(\frac{19}{27}, \frac{20}{27}\right)\cup \left(\frac{25}{27}, \frac{26}{27}\right) \right\}\\ &=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27}, 1\right]\\ \vdots \end{align*}$$

The ultimate goal is to show that the cantor set is compact. I know that it is bounded between $[0,1]$, but I need to use an induction proof to show that it is closed.

I have only gotten as far as the base case. The complement for the interval $[0,1]$ is the union of open sets $(-\infty,0)\cup (1,\infty)$. Hence the interval $[0,1]$ is closed.

How would one set up this induction?

Answer

If you look at the actual construction of $S_{k+1}$ from $S_k$, you’ll see that $S_{k+1}=S_k\setminus U_k$, where $U_k$ is the union of a certain finite collection of open intervals in $[0,1]$. Thus, $S_{k+1}=S_k\cap\Big([0,1]\setminus U_k\Big)$, the intersection of two closed sets, and so must itself be closed. That’s the induction step in outline right there; all you have to do is fill in a few details about $U_k$.

Once you’ve shown by induction that all of the sets $S_k$ are closed, you just use the fact that any intersection of closed sets is automatically a closed set.