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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?

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    You said you were going to define $S_k$ "in the following manner" but never said how...2012-11-10

3 Answers 3

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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.

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    @Joakim: No; why would you think so? At stage $n$ you’re removing all of the intervals $\left(\frac{2k+1}{3^n},\frac{2k+2}{3^n}\right)$ for $k=0,\dots,\frac23(3^{n-1}-1)$ no matter whether $n$ is even or odd.2012-11-11
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$S(k)$ is closed for all k and the intersection of a family of closed sets yields a closed set.

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If you show, by induction, or by other means, that $S_k$ is closed for all $k$ then $C$ is the intersection of closed sets, and therefore closed as well.

You simply have to show (perhaps by induction, perhaps... by definition) that $S_k$ being a finite union of closed intervals is closed.

If you were given the axioms of topology this is trivial, if you were not given such axioms then this may take a bit more work. However you might want to prove the following claim:

Suppose $[a_1,b_1],\ldots[a_n,b_n]$ are closed intervals, then $\bigcup_{i=1}^n[a_1,b_1]$ is closed.

(For your case it is sufficient to show that for disjoint intervals, of course)

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    @Jo$a$kim: I see now. You added information to the question. The work may depend on how you define a closed set. But it is possi$b$le that what I wrote above is something you will have to do with an indu$c$tion.2012-11-10