Here is Prob. 6, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Let $A_0$ be the closed interval $[0, 1]$ in $\mathbb{R}$. Let $A_1$ be the set obtained from $A_0$ by deleting its "middle third" $\left( \frac{1}{3}, \frac{2}{3} \right)$. Let $A_2$ be the set obtained from $A_1$ by deleting its "middle thirds" $\left( \frac{1}{9}, \frac{2}{9} \right)$ and $\left( \frac{7}{9}, \frac{8}{9} \right)$. In general, define $A_n$ by the equation $$ A_n = A_{n-1} - \bigcup_{k=0}^\infty \left( \frac{1+3k}{3^n}, \frac{2+3k}{3^n} \right). $$ The intersection $$ C = \bigcap_{n \in \mathbb{Z}_+ } A_n $$ is called the Cantor set; it is a subspace of $[0, 1]$.
(a) Show that $C$ is totally disconnected.
(b) Show that $C$ is compact.
(c) Show that each set $A_n$ is a union of finitely many disjoint closed intervals of length $1/3^n$; and show that the end points of these intervals lie in $C$.
(d) Show that $C$ has no isolated points.
(e) Conclude that $C$ is uncountable.
My Attempt:
First of all, we have $$ \begin{align} A_0 &= [0, 1] \\ &= \left[ 0, \frac{1}{3^0} \right], \\ A_1 &= A_0 \setminus \left( \frac{1}{3}, \frac{2}{3} \right) = [0, 1] \setminus \left( \frac{1}{3}, \frac{2}{3} \right) = \left[ 0, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, 1 \right] \\ &= \left[ 0, \frac{1}{3^1} \right] \cup \left[ \frac{2}{3^1}, 1 \right], \\ A_2 &= A_1 \setminus \big( \left( \frac{1}{9}, \frac{2}{9} \right) \cup \left( \frac{7}{9}, \frac{8}{9} \right) \big) = \big( \left[ 0, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, 1 \right] \big) \setminus \big( \left( \frac{1}{9}, \frac{2}{9} \right) \cup \left( \frac{7}{9}, \frac{8}{9} \right) \big) \\ &= \left[ 0, \frac{1}{9} \right] \cup \left[ \frac{2}{9}, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, \frac{7}{9} \right] \cup \left[ \frac{8}{9}, 1 \right] \\ &= \left[ 0, \frac{1}{3^2} \right] \cup \left[ \frac{2}{3^2}, \frac{3}{3^2} \right] \cup \left[ \frac{6}{3^2}, \frac{7}{3^2} \right] \cup \left[ \frac{8}{3^2}, 1 \right], \\ A_3 &= A_2 \setminus \big( \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) \big) \\ &= \big( \left[ 0, \frac{1}{9} \right] \cup \left[ \frac{2}{9}, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, \frac{7}{9} \right] \cup \left[ \frac{8}{9}, 1 \right] \big) \setminus \big( \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) \big) \\ &= \left[0, \frac{1}{27} \right] \cup \left[ \frac{2}{27}, \frac{1}{9} \right] \cup \left[ \frac{2}{9}, \frac{7}{27} \right] \cup \left[ \frac{8}{27}, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, \frac{19}{27} \right] \cup \left[ \frac{20}{27}, \frac{7}{9} \right] \cup \left[ \frac{8}{9}, \frac{25}{27} \right] \cup \left[ \frac{26}{27}, 1 \right] \\ &= \left[0, \frac{1}{3^3} \right] \cup \left[ \frac{2}{3^3}, \frac{3}{3^3} \right] \cup \left[ \frac{6}{3^3}, \frac{7}{3^3} \right] \cup \left[ \frac{8}{3^3}, \frac{9}{3^3} \right] \cup \left[ \frac{18}{3^3}, \frac{19}{3^3} \right] \cup \left[ \frac{20}{3^3}, \frac{21}{3^3} \right] \cup \left[ \frac{24}{3^3}, \frac{25}{3^3} \right] \cup \left[ \frac{26}{3^3}, 1 \right], \\ & \vdots \\ A_n &= \ldots. \end{align} $$
Can we write down a general formula for $A_n$?
We note that each $A_n$ is a union of $2^n$ disjoint closed intervals each of which has length $\frac{1}{3^n}$.
Part (a)
Let $\{ x, y \}$ be any two-point subset of $C$. We assume without loss of generality that $x < y$. Let $n$ be any natural number such that $$ 3^n > \frac{1}{y-x}.$$ Then we have $$ 0 < \frac{1}{3^n} < y-x. $$ Thus $x$ and $y$ belong to two distinct closed intervals, which are disjoint, that constitute $A_n$. Let $z$ be any number such that $z \not\in A_n$ and such that $x < z < y$. Then the following constitutes a separation of $\{ x, y\}$: $\{ x, y \} \cap (-\infty, z)$ and $\{ x, y \} \cap (z, +\infty)$.
Thus every two-point subset of $C$ is disconnected.
Next, suppose $S$ is any subset of $C$ such that $S$ has more than one point. Let $x, y \in S$ such that $x < y$. As before let us choose a point $z$ such that $z \not\in C$ and $x < z < y$. Then the sets $S \cap (-\infty, z)$ and $S \cap (z, +\infty)$ constitute a separation of $S$.
Thus any subset $S$ of $C$ having more than one point is disconnected.
Therefore the Cantor set $C$ is totally disconnected.
Is this proof correct and clear enough? If so, does it have sufficient detail?
Part (b)
We note that each set $A_n$ is a union of finitely many closed intervals and is therefore closed in $\mathbb{R}$. Moreover since each $A_n \subset [0, 1]$, so $A_n$ is also closed in $[0, 1]$. Thus the Cantor set $C$, being an intersection of the closed sets $A_n$, is also closed in $[0, 1]$.
Finally, as $[0, 1]$ is compact (as a subspace of $\mathbb{R}$ with standard topology) and as $C$ is closed in $[0, 1]$, so the Cantor set $C$ is also compact.
Is this proof correct? If so, is it clear and detailed enough?
Part (c)
We note that $A_0 = [0, 1]$ so that $A_0$ is (a union of) a single closed interval of length $1 = 1/3^0$. Both the endpoints are in $A_0$ of course.
Also $A_1 = \left[ 0, \frac{1}{3} \right] \cup \left[ \frac{2}{3}, 1 \right] $ so that $A_1$ is the union of two disjoint closed intervals, each of length $1/3^1$. Again all the endpoints, namely $0, 1/3, 2/3$, and $1$, are in $A_1$.
Now suppose that $A_n$ is a union of finitely many ($2^n$ to be precise) disjoint closed intervals, each of length $1/3^n$, and thus we find that all the endpoints of these intervals are in $A_n$.
Then we note that $A_{n+1}$ is obtained from $A_n$ by deleting the "middle third" open interval from each of the closed intervals that $A_n$ is composed of. In this way each of the $2^n$ closed intervals that $A_n$ is composed of yields two smaller disjoint closed intervals, and each of these new intervals has length equal to one-third the length of its parent interval.
Thus $A_{n+1}$ is a union of $2 \times 2^n = 2^{n+1}$, and thus finitely many, disjoint closed intervals, each having length $\frac{1}{3} \times \frac{1}{3^n} = \frac{1}{3^{n+1}}$.
Moreover, if $[a, b]$ is any closed interval contained in $A_n$, where $a$ and $b$ are real numbers such that $$ 0 \leq a < b \leq 1, $$ then after deleting the "middle third" open interval, which is $\left( a + \frac{b-a}{3}, a + 2\frac{b-a}{3} \right)$, we are left with $$ [a, b] \setminus \left( a + \frac{b-a}{3}, a + 2\frac{b-a}{3} \right) = \left[ a, a + \frac{b-a}{3} \right] \cup \left[ a + 2\frac{b-a}{3}, b\right], $$ which is a union of two disjoint closed intervals that form part of $A_{n+1}$ and thus all the endpoints of these intervals are in $A_{n+1}$.
In short, in forming $A_{n+1}$ from $A_n$, the endpoints of all the disjoint closed intervals that $A_n$ is composed of are retained.
Furthermore, since $$A_0 \supset A_1 \supset A_2 \supset \cdots, $$ we note that all the endpoints of the disjoint closed intervals contained in any set $A_n$ are in all the other sets as well.
Hence by induction we have shown that, each set $A_n$ is a union of finitely many (i.e. $2^n$) disjoint closed intervals of length $1/3^n$ and also that the endpoints of these intervals are retained in each set $A_n$ and are therefore in $C$.
Is this proof correct in each and every detail? Or are there issues?
Part (d)
Let $x$ be any point of $C$. Then $x$ is in each set $A_n$. Let $I_n$ be the closed interval in $A_n$ containing $x$.
Let $\varepsilon > 0$ be given. Let $N$ be any natural number such that $1/3^N < \varepsilon$. Let $a_N$ be endpoint of $I_N$ such that $a_N \neq x$. Then we note that $a_N \in C$ and also that $$ 0 < \left\lvert a_N - x \right\rvert \leq \frac{1}{3^N} < \varepsilon, $$ which implies that $x$ is a limit point of $C$.
Thus every point of $C$ is a limit point of $C$. Or in other words, no singleton subset of $C$ is open in $C$. Hence $C$ is no isolated points.
Is this proof correct?
Part (e)
Since the closed interval $[0, 1]$ is a subspace of the Hausdorff space $\mathbb{R}$, so $[0, 1]$ is also Hausdorff. And, since $C$ is a subspace of $[0, 1]$ and since $[0, 1]$ is Hausdorff, so is $C$.
By Parts (b) and (d) we also know that $C$ is compact and $C$ has no isolated points.
Thus $C$ is a non-empty compact Hausdorff space having no isolated points. Therefore by Theorem 27.7 in Munkres $C$ is uncountable.
Is my proof correct?
Is each one of my proofs correct? If so, is each one of these proofs rigorous enough? Is each proof also clear and detailed enough for a beginner to make sense of?
Or, are there any issues of accuracy of logic, clarity, or rigor in any of these proofs?
I believe I do find a general formula for $A_n$ by observing:
Define $J_0=\{0\}$ and $J_n=\bigcup\limits_{k=0}^{n-1}\left\{\sum\limits_{i=k}^{n-1} 3^i\right\}$ for all $n\in \mathbb{Z}_+$, put $Z_n=\bigcup\limits_{m=0}^n J_m$. Then $$A_n=\bigcup_{k\in Z_n}\left[\frac{2k}{3^n},\frac{2k+1}{3^n}\right].$$
It fits well for $n=0,1,2,3$. However, I find difficulty proving this. I'd be grateful if anybody could provide some hints for it.