# Cantor's Set Has No Intervals

I'm asked to prove that the Cantor set has no intervals. This is what I've got:

Let $(a,b)\subseteq [0,1]$, where $a<b$. Let $T=\{-\log_3(b-a)<n:n\in\mathbb{N}\}$. Now if $(a,b)\subseteq[0,1]$, then $b-a<1$, and so $-\log_3(b-a)>0$. This would then mean that $-\log_3(b-a)\in\mathbb{R}$, and so by the Archimidean property there exists some $\alpha\in T$ such that $a\leq \beta$ for all $\beta\in T$. Hence $$-\log_3(b-a)<\alpha,$$ so we have know that $$3^{-(-\log_3(b-a))}>3^{-\alpha}.$$ Of course the LHS here reduces the inequality as $$b-a=\lvert b-a \rvert < 3^{-\alpha}.$$ Now $A_{\alpha}$ is the union of subsets of $[0,1]$ having length $3^{-\alpha}$, so $(a,b)\not\subseteq A_{\alpha}$; hence $(a,b)\not\subseteq C$, as was to be shown.

I'm wondering what your opinion on the proof is. Good? Bad? Incorrect? Throw it away! There's a faster way! You know, all of this.

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What is $A_\alpha$ in your proof? – Mauricio G Tec Sep 3 '13 at 20:44
Union of subsets of $[0,1]$ having length $3^{-\alpha}$. – Trancot Sep 3 '13 at 20:49
I believe the set you mean as $T$ is $\{n\in\mathbb N\mid -\log_3(b-a)<n\}$. Read this as "the set of $n$ in $\mathbb N$ such that ..." The set you wrote reads as "the set of $-\log_3(b-a)$ that are less than $n$ where $n$ varies over $\mathbb N$", which is most likely not the set you meant. – Andrés E. Caicedo Sep 3 '13 at 20:54
Union of which subsets of $[0,1]$ having length $3^{-\alpha}$? – Trevor Wilson Sep 3 '13 at 20:55

You’ve a notational problem with your definition of $T$: it should read

$$T=\{n\in\Bbb N:-\log_3(b-a)<n\}\;.$$

The generic object goes before the colon, and the defining condition goes after it. The next part of the argument is a bit confused. The Archimedean property says that $T\ne\varnothing$, but you need the well-ordering principle for $\Bbb N$ to conclude that $T$ has a least element $\alpha$. Now you have the inequality $-\log_3(b-a)<\alpha$, so $\log_3(b-a)>-\alpha$, and $b-a>3^{-\alpha}$, but in the next line you somehow reversed the inequality; I’m going to assume that this was just a typo, since you seem to be using the correct inequality to get your conclusion.

I say seem to because your next step isn’t entirely clear. I think that you want $A_\alpha$ to be the the set $C_\alpha$ in this construction of the middle-thirds Cantor set: the stage at which you have a union of pairwise disjoint closed intervals of length $3^{-\alpha}$. Then you’re arguing that since $b-a>3^{-\alpha}$, $(a,b)$ cannot be a subset of any of those intervals and therefore cannot be a subset of $A_\alpha$, let alone of the Cantor set itself. If that’s what you intend, the argument is basically correct.

There are other ways to prove the result. If you’ve already proved that the middle-thirds Cantor set $C$ contains precisely those real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, you can easily prove that $[0,1]\setminus C$ is dense in $C$. Suppose that $x,y\in C$, where

$$x=\sum_{k\ge 1}\frac{a_k}{3^k}\quad\text{and}\quad y=\sum_{k\ge 1}\frac{b_k}{3^k}\;,$$

where all $a_k,b_k\in\{0,2\}$. Let $n\in\Bbb Z^+$ be minimal such that $a_n\ne b_n$, and without loss of generality assume that $a_n=0$ and $b_n=2$. Define $c_k=a_k=b_k$ for $k=1,\dots,n-1$ and $c_k=1$ for $k\ge n$, and let $$a=\sum_{k\ge 1}\frac{c_k}{3^k}\;;$$ then $x<z<y$, and $z\notin C$.

Closer in spirit to your own argument, you can calculate the total length of all of the open intervals that are removed in the construction:

$$\frac13+2\cdot\frac19+4\cdot\frac1{27}+\ldots=\frac13\sum_{k\ge 0}\left(\frac23\right)^k=\frac13\cdot\frac1{1-\frac23}=1\;.$$

Assuming that $C$ contains a non-empty open interval then leads to an immediate contradiction.

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It would be easier to read the proof if you start with an overview of the strategy, such as "We will find a level of the construction of the Cantor set that does not contain the interval." The definition of $T$ would make more sense if written as $T=\{n\in\mathbb{N}:-\log_3(b-a)<n\}$. When you invoke the Archimedean property, you're really proving that $T$ is not empty. Then, yes, it's true that a nonempty subset of $\mathbb N$ has a least element $\alpha$. However, you don't actually need $\alpha$ to be the least element of $\mathbb N$, so you should omit that condition. Just let $\alpha$ be any member of $T$.

I would say that at some point in an analysis course, you no longer have to mention the Archimedean property in every proof involving the real numbers. It would be quicker to say something like this:

Any interval contained in the $n$th level of the construction $A_n$ has length $\leq 3^{-n}$, so any interval contained in the Cantor set $\bigcap A_n$ has length $\leq \inf 3^{-n}=0$.

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Here is a slightly faster, beautiful but trickier way (Recall that elements in $C$ are characterized by the fact that they can be written in base 3 using only the digits 0 and 2):

Suppose $C$ does contains an interval $(a,b)$. Then $(a,b)$ contains a number $x$ that can be written in base-3 as $x=.d_1d_2...$ with all the $d_i\in\left\{0,2\right\}$. Note that if two numbers written in ternary expansion have the first $k$th digits the same then they can differ in at most $1/3^k$. Set $\delta=\min\{x-a,b-x\}$ and choose $k$ big enough such that $1/3^k<\delta$ (if extreme formality is demanded use the Archimedean principle here). Define now $y=.d_1d_2...d_k1111111...$. By the preceding remark $|x-y|<1/3^k<\delta$ and therefore $y\in(a,b)$. This is a contradiction though, since this number cannot be written using only $0$'s and $2$'s and so $y\notin C$.

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