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(i) Define the Cantor set.
E_{n}=\bigcup_{k=1 \atop 3|k, 3|(k-2)}^{3^{n}-1}\left[\frac{k}{3^{n}}, \frac{k+1}{3^{n}}\right]
The Cantor set, E, is defined to be
E=\bigcap_{n=1}^{\infty} E_{n}
(ii) Show that the Cantor set is perfect.
in Mathematics by Bronze Status (8,688 points) | 10 views

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Proof. Each $E_{n}$ is a finite union of closed intervals and hence closed. $E$ is an intersection of closed sets and hence is also closed. If $x \in E$, then $x \in E_{n}$ for all $n$. Therefore, let $I_{n}$ be the closed interval of $E_{n}$ such that $x \in I_{n}$ and let $x_{n}$ be the endpoint of $I_{n}$ with $x_{n} \neq x$. Note that $I_{n}$ is an interval of length $3^{-n}$. Hence we have
We certainly have $x_{n} \neq x$ and the size of their difference can be make smaller than any $\epsilon>0$ by taking $n$ large enough, larger than $-\log _{3}\left(\min \left\{\varepsilon, \frac{1}{2}\right\}\right)$.
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