You can see the original text that i thought AC is used here;

From Walter Rudin: Principles of Mathematical Analysis, 3rd ed., ISBN 0-07-054235-X, p.41-42.

2.44 The Cantor set The set which we are now going to construct shows that there exist perfect sets in $R^1$ 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 really didn't like the definition of Cantor set in my book (It defines cantor set by using AC$_\omega$), so i tried to construct it in more constructive way. (That is, without AC)

Let $E_n = [0,1]\setminus \bigcup_{k\in 3^n}(3k+1/3^{n+1}, 3k+2/3^{n+1})$.

(Exponentiation here is an ordinal exponentiation)

Let $C=\bigcap_{n\in \omega} E_n$.

I proved that $C$ is nonempty and compact, and $C$ contains no segment.

Now, I'm trying to prove that $C$ is perfect but there's a problem.

This is a lemma I made to prove this; Let $T_n = \{3k/3^{n+1}, 3k+1/3^{n+1}, 3k+2/3^{n+1}, 3k+3/3^{n+1} \in [0,1] | k\in 3^n \}$.

'For every $n,m\in \omega$, if $n<m$, then $T_n\subset T_m$.'

Now, fix $x\in C$ and $0<r\in \mathbb{R}$. Here, i have proved that there exists $m\in \omega$ such that $x$ is in some interval $I$ in $E_m$ and the $I\subset B(r,x)$. Now let $x'$ be an endpoint such that $x'≠x$.

By the lemma above, $x'\in E_n$ for every $m≦n$. However, i don't know how to prove this for $m>n$.

I tried it and proved that 'If $m>n$ and intersection of an interval $I_n$ in $E_n$ and an interval $I_m$ in $E_m$ is nonempty, the intersection is a set of exactly one endpoint of $I_m$ or $I_m \subset I_n$. (If one holds, the other does not hold)' too. Here, I have no idea how to derive a contradiction that 'if $x\in I_m \cap I_n$, then $I_m \cap I_n$ is NOT a singleton' ($x\in C$)


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    $\begingroup$ Try to see the fact that $C$ is the set of reals in $(0,1)$ having only $0$s and $1$s in their ternary expansion. Now, prove that there are no isolated points, and thus prove that $C$ is perfect. $\endgroup$ – Rijul Saini Aug 1 '12 at 3:47
  • $\begingroup$ I think you mean only zeros and twos. $\endgroup$ – Stefan Geschke Aug 1 '12 at 4:56
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    $\begingroup$ I wonder what the definition with DC looks like. I can't think of one. $\endgroup$ – t.b. Aug 1 '12 at 6:12
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    $\begingroup$ yeah, it would be interesting to ee that. @Katlus: please add a complete reference to the book: saying «my book» does not help in finding it :D $\endgroup$ – Mariano Suárez-Álvarez Aug 1 '12 at 6:19
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    $\begingroup$ Mariano's question still stands: what is your book? And by "defines the Cantor set by $\mathrm{AC}_\omega$" do you mean the definition of the Cantor set as $\{0,1\}^\mathbb{N}$? No $\mathrm{AC}_\omega$ here. This is nothing but the power set $P(\mathbb{N})$, which you know has cardinality $\mathfrak{c}$ with or without choice. $\endgroup$ – t.b. Aug 1 '12 at 7:33

Converting my above comment into an answer:

While Rudin may be accused on hand-waving, I'm pretty sure that his basic idea uses no Choice principles. Note that once $E_n$ is given, then the next set $E_{n+1}$ is determined, and not chosen out of some collection of possibilities (or, to hedge, it is chosen out of a collection of exactly one possibility, but no Choice is required to do this). This is a recursive construction (much like the definition of ordinal arithmetic is recursive).


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