TITLE: Cantor's Set Has No Intervals
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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.