Exercise 3.3.7

Question

As some more evidence of the surprising nature of the Cantor set, follow these steps to show that the sum $C+C=\{x+y: x, y \in C\}$ is equal to the closed interval $[0,2]$. (Keep in mind that $C$ has zero length and contains no intervals.)

Because $C \subseteq[0,1], C+C \subseteq[0,2]$, so we only need to prove the reverse inclusion $[0,2] \subseteq\{x+y: x, y \in C\}$. Thus, given $s \in[0,2]$, we must find two elements $x, y \in C$ satisfying $x+y=s$
  \begin{enumerate}[label=(\alph*)] 
  \item Show that there exist $x_{1}, y_{1} \in C_{1}$ for which $x_{1}+y_{1}=s$. Show in general that, for an arbitrary $n \in \mathbf{N}$, we can always find $x_{n}, y_{n} \in C_{n}$ for which $x_{n}+y_{n}=s .$
  \item Keeping in mind that the sequences $\left(x_{n}\right)$ and $\left(y_{n}\right)$ do not necessarily converge, show how they can nevertheless be used to produce the desired $x$ and $y$ in $C$ satisfying $x+y=s$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Recall the definition for sets $A$ and $B$, $A + B = \{x + y: x \in A, \ y \in B\}$. Note that $[a, b] + [c, d] = [a + c, b+d]$.
Recall $C_1 = [0, 1/3] \cup [2/3, 1]$; we have
    $$
        \begin{aligned}
        C_1 + C_1 &= ([0,1/3] + [0,1/3]) \cup ([2/3,1] + [2/3,1]) \cup ([0,1/3] + [2/3,1]) \
        &= [0, 2/3] \cup [4/3, 2] \cup [2/3,4/3] \
        &= [0, 2]
        \end{aligned}
    $$

We prove the general case through induction; the base case has been demonstrated above. Define $A + r = \{x+r : x \in A\}$, $r \cdot A = \{rx : x \in A\}$ for $r \in \mathbf{R}$. Note that $[a,b] + c = [a + c, b+c]$ and $r \cdot (A + B) = r \cdot A + r \cdot B$.

The inductive hypothesis is that $C_n + C_n = [0, 2]$. We can complete the inductive step by noticing that if we scale up $C_{n+1}$ a factor of 3, we simply get two copies of $C_n$, with one being offset 2 away; this ultimately makes it easy to express $C_{n+1}$ in terms of $C_n$. Let $C_{n+1}' = C_n \cup (C_n + 2) = 3 \cdot C_{n + 1}$ be this upscaled $C_{n+1}$. Then
    $$ \begin{aligned}
        C_{n+1}' + C_{n+1}' &= (C_n + C_n) \cup (C_n + (C_n + 2)) \cup ((C_n + 2) + (C_n + 2)) \
        &= [0, 2] \cup ([0, 2] + 2) \cup ([0, 2] + 4)\
        &= [0, 6]
        \end{aligned}
$$
    $$\frac{1}{3} \cdot (C_{n+1}' + C_{n+1}')  = C_{n+1} + C_{n+1} = \frac{1}{3}\cdot [0,6] = [0,2] $$
    and the inductive step is complete.

\item Since $C$ is compact, there exists a subsequence $(x_{n_k}) \to x$ with $x \in C$. Now since $x_{n_k} + y_{n_k} = s$ for all $k$, we have $\lim y_{n_k} = \lim s - x_{n_k} = s - x$. Now since each $y_{n_k} \in C$ the limit $y = s - x \in C$ as well, thus we have found $x,y\in C$ with $x+y = s$.
   \end{enumerate}

Exercise 3.3.7

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Exercise 3.3.7

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