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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.

Quasar
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    Closely related: https://math.stackexchange.com/questions/530512/show-that-the-cantor-set-is-perfect-i-e-c-c?rq=1, https://math.stackexchange.com/questions/1090736/a-proof-that-the-cantor-set-is-perfect?rq=1, https://math.stackexchange.com/questions/2477793/cantor-set-nowhere-dense-but-still-perfect?rq=1 – Lee Mosher Jun 12 '21 at 15:49
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    Your idea is completely correct. Proof-wise you have made a few leaps, namely "In general, we can pick a number $\displaystyle x_{n}$ ..." which was the entire question, but this is understandable as the Cantor set is difficult to notate. – aristotlefromgreece Jun 12 '21 at 16:00
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    As a side note, this is also intuitively obvious if you know/can use the fact that the elements of the cantor set are exactly those whose base 3 representation uses nothing but 0's and 2's. From that, we can always find elements within any epsilon by shifting a 0 to a 2 or visa versa sufficiently down the road – Alan Jun 12 '21 at 16:40
  • @Alan, thanks! Yes, if I understand correctly, any number in the Cantor set can be uniquely represented by an infinite string which is a base-3 number made of 0s and 2s, where $0$ implies taking the left branch, whereas $2$ implies picking the right branch. – Quasar Jun 12 '21 at 17:08
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    @quasar Exactly, so given any one point and any epsilon, you can find another point within that just by moving down the base 3 string enough spaces – Alan Jun 12 '21 at 18:54

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