I am self-learning Real Analysis. Stephen Abbott's book leaves the proof of the fact that the Cantor set $C$ has no isolated points as an exercise. I'd like someone to verify, if my construction is correct, or point out any fallacies.
[Abbott, 3.4.3] Review the portion of the proof given in Example 3.4.2 to show that the Cantor set is perfect and follow these steps to complete the argument.
(a) Because $\displaystyle x\in C_{1}$, argue that there exists an $\displaystyle x_{1} \in C\cap C_{1}$ with $\displaystyle x_{1} \neq x$ satisfying $\displaystyle | x-x_{1}| \leq 1/3$.
(b) Finish the proof by showing that for each $\displaystyle n\in \mathbf{N}$, there exists $\displaystyle n\in \mathbf{N}$, there exists $\displaystyle x_{n} \in C\cap C_{n}$, different from $\displaystyle x$, satisfying $\displaystyle | x-x_{n}| \leq 1/3^{n}$.
Proof.
Let $\displaystyle x\in C$ be an arbitrary element. Since, $\displaystyle C=\bigcap _{n=0}^{\infty } C_{n}$, $\displaystyle x\in C_{1}$.
Exactly, one of the below two possibilities must be true.
(1) $\displaystyle x\in \left[ 0,\frac{1}{3}\right]$
(2) $\displaystyle x\in \left[\frac{2}{3} ,1\right]$
In either case, we can pick a number $\displaystyle x_{1}$ to be the left or right end-point of one of the two intervals in $\displaystyle C_{1}$ different from $\displaystyle x$. Since the end-points of $\displaystyle C_{n}$ are contained in $\displaystyle C$, we can pick an $\displaystyle x_{1} \in C\cap C_{1}$, such that $\displaystyle x_{1} \neq x$ and $\displaystyle | x_{1} -x| \leq \frac{1}{3}$.
Similarly, $\displaystyle x\in C_{2}$. If $\displaystyle x\in \left[ 0,\frac{1}{3}\right]$, then we have a dichotomy. Either $\displaystyle x\in \left[ 0,\frac{1}{9}\right]$ or $\displaystyle x\in \left[\frac{2}{9} ,\frac{3}{9}\right]$. If $\displaystyle x\in \left[\frac{2}{3} ,1\right]$, then either $\displaystyle x\in \left[\frac{6}{9} ,\frac{7}{9}\right]$ or $\displaystyle x\in \left[\frac{8}{9} ,1\right]$. So, we can pick $\displaystyle x_{2}$ to be the left or right end-point of one of the two intervals in $\displaystyle C_{2}$ different from $\displaystyle x$. Since the end-points of $\displaystyle C_{n}$ are contained in $\displaystyle C$, we can pick an $\displaystyle x_{2} \in C\cap C_{2}$, such that $\displaystyle x_{2} \neq x$ and $\displaystyle | x_{2} -x| \leq \frac{1}{9}$.
In general, we can pick a number $\displaystyle x_{n}$ to be the left or right end-point of one of two intervals in $\displaystyle C_{n}$, such that $\displaystyle x_{n} \neq x$ and $\displaystyle | x_{n} -x| \leq \frac{1}{3^{n}}$.
Now, consider the sequence $\displaystyle ( x_{n}) =( x_{1} ,x_{2} ,\dotsc )$. Given any arbitrary $\displaystyle \epsilon >0$, pick $\displaystyle n_{0}$ such that
\begin{equation*} \frac{1}{3^{n_{0}}} < \epsilon \quad \text{ or } \quad n_{0} >\left(\log\frac{1}{\epsilon }\right) /(\log 3) \end{equation*} then for $\displaystyle n\geq n_{0}$, we have $\displaystyle | x_{n} -x| < \epsilon $. So, $\displaystyle ( x_{n})\rightarrow x$.
Thus, $\displaystyle C$ is a perfect set.