Exercise 3.4.4

Question

Repeat the Cantor construction from Section $3.1$ starting with the interval $[0,1]$. This time, however, remove the open middle \emph{fourth} from each component.
  \begin{enumerate}[label=(\alph*)] 
  \item Is the resulting set compact? Perfect?
  \item Using the algorithms from Section 3.1, compute the length and dimension of this Cantor-like set.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item The proofs that the Cantor set is compact and perfect can be copied and applied nearly word for word here. The Cantor-like set is obviously bounded, and it is closed because it is the intersection of countably many closed sets (see Exercise 3.2.6e); therefore it must be compact. Using the same strategy as Exercise 3.4.3, for any $x$ in the Cantor-like set, we can find a sequence $(x_n)$ where $x_n \neq x$ but $|x_n - x| \leq (3/8)^n $.
  \item The sum of the lengths of the removed segments is
  $$\frac{1}{4} + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right) + \cdots + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{n - 1} + \cdots = 1$$
and thus the Cantor-like set has zero length.

Magnifying the Cantor-like set by a factor of $8/3$ leaves us with two copies of the set, hence the dimension $d = \log 2 / \log (8/3) \approx 0.707$.
   \end{enumerate}

Exercise 3.4.4

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Exercise 3.4.4

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