In Chapter 2 of Rudin's Priniciples of Mathematical Analysis, Rudin takes the Cantor set as an example of a perfect set in $\mathbb{R}^1$ which contains no segment. Here's the construction of the Cantor set and the proof:
2.44 The Cantor set The set which we are now going to construct shows
that there exist perfect sets in $\mathbb{R}^1$ which contain no segment.Let $E_0$ be the interval $[0, 1]$. Remove the segment $(\frac13,\frac23)$, and let $E_1$ be
the union of the intervals
$$[0,\frac13], [\frac23,1].$$
Remove the middle thirds of these intervals, and let $E_2$ be the union of the
intervals
$$[0,\frac19], [\frac29,\frac39], [\frac69,\frac79],[\frac89,1]$$
Continuing in this way, we obtain a sequence of compact sets $E_n$, such that
(a) $E_1\supset E_2 \supset E_3 \dots $;
(b) $E_n$ is the union of $2^n$ intervals, each of length $3^{-n}$.The set
$$P=\bigcap_{n=1}^\infty E_n$$
is called the Cantor set. $P$ is clearly compact, and Theorem 2.36 shows that $P$
is not empty.No segment of the form
$$\left(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\right)\tag{24},$$
where $k$ and $m$ are positive integers, has a point in common with $P$.
Since every segment $(\alpha,\beta)$ contains a segment of the form (24), if
$$3^{-m}<\frac{\beta-\alpha}6,$$
$P$ contains no segment.To show that $P$ is perfect, it is enough to show that $P$ contains no isolated
point. Let $x \in P$, and let $S$ be any segment containing $x$. Let $I_n$ be that interval
of $E_n$ which contains $x$. Choose $n$ large enough, so that $I_n\subset S$. Let $x_n$ be an
endpoint of $I_n$, such that $x_n\ne x$.It follows from the construction of $P$ that $x_n\in P$. Hence $x$ is a limit point
of $P$, and $P$ is perfect.One of the most interesting properties of the Cantor set is that it provides
us with an example of an uncountable set of measure zero (the concept of
measure will be discussed in Chap. 11).
I can follow the proof with some effort, but in the end of this section Rudin claims that the Cantor set is an example of an uncountable set of measure zero. How can the Cantor set be uncountable? Corollary of Theorem 2.13 shows the set of all rational numbers is countable. Theorem 2.8 shows that every infinite subset of a countable set is countable. The elements in the Cantor set are the end points of all the intervals in $E_n$, it follows from the construction of the Cantor set that these end points are all rational numbers. Hence $P$ is a subset of the rational numbers and countable. Is there anything wrong with my reasoning here?
Best Answer
"The elements in the Cantor set are the end points of all the intervals in $E_n$..." This is your mistake. This isn't true. In fact, written in ternary expansion, the elements of the Cantor set are precisely those elements in $[0,1]$ with a ternary expansion consisting of $0$'s and $2$'s (where we note $0.01=0.00\bar{2}\in\mathcal{C}$, but $0.0101\notin\mathcal{C}$, for example). Using this fact, it isn't hard to show that $\frac{1}{4}\in\mathcal{C}$ but $1/4$ is not an endpoint of any interval.