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page 49-50 Exercise 3.19

Let $A=\{0,2\}$ and $C$ be the Cantor Set. Define $x(\alpha) = \sum_{n=1}^\infty (\alpha_n / {3^n})$ for all $\alpha \in A^{\mathbb{N}}$.

Then $x$ is a well defined function.

I think the argument in the link assumed, without any notice, the existence of a sequence $\beta$ such that $x(\beta) = z$, for each $z\in C$.

Am I correct? Hence, the argument proves only ${ran} x \subset C$.

How do i prove that there exists a such sequence for each $z\in C$?

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I am not sure if it's widely used to denote range set as $ran$. – Katlus Oct 7 '12 at 18:06

Consider the ternary (base 3) representation of numbers in the interval $[0, 1]$. Just like how $1$ can also be represented as $0.999...$ in decimal, $1/3$ can be represented as $0.1$ or $0.0222...$ in base 3. Also, $2/3$ can be represented as $0.2$ or $0.1222...$ in base 3. In between $1/3$ and $2/3$, all numbers start with $0.1...$ in base 3. Therefore, when we remove the middle third from $[0, 1]$, we remove exactly the numbers that start with $0.1...$ in base 3. For interval boundaries, we have 2 choices for base 3 representations, so we use the ones that don't contain $1$.

As we continue removing intervals to build the Cantor set, at each step $n$, we remove exactly the numbers that have $1$ at position $n$ in their ternary representation. The Cantor set consists of exactly the numbers that don't have $1$ anywhere in their ternary representation. Thus, there is a 1:1 mapping between the cantor set and $\{0, 2\}^\mathbb{N}$. Namely, each number in the Cantor set has a sequence of $\{0, 2\}$ as its ternary representation.

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I have a problem. How can I express this in meta? I can't use recursion theorem since i don't know how to express this in $n$th term. Here's what i tried. Fix $x$ in a Cantor Set. Let $A=${$n\in \mathbb{N}$|$x$ has a ternary representation such that for all $m≦n$, $m$th term is not $1$}. Here, recursion works, but i can't prove "There exists a ternary representation of $x$ such that every term is not $1$".( Assume negation of this sentence. If i choose one representation, i have to change the representation to fit, then changed one has another $m\in\mathbb{N}$ such that $m$th term is $1$. – Katlus Oct 7 '12 at 20:57
You can use induction. All strings from $\{0,2\}^m\{0,1,2\}^{\mathbb{N}}$ represent what you get after $m$ steps of "removing the middl third". Induction then gives the proof. – yo' Oct 7 '12 at 21:23
Yep, use induction. I've shown the base case above. Proving the case of $m+1$ after $m$ steps is done in a very similar manner. – Ayman Hourieh Oct 7 '12 at 21:27
I still don't get it.. What I'm talking is not really about induction. By induction, "For each $n\in \mathbb{N}$, there exists a ternary representation such that for all $m≦n$, $m$th term is not $1$.' is true. (This is exactly what tohecz says) However, I cannot derive "There exists a ternary representation such that every term is not $1$' from the first sentence.. – Katlus Oct 7 '12 at 21:59
Ternary representation is, in other words, a function.(Actually it's not, but if we only allow one representation for each $x$,then i would be a function.) Since this function is not 'fixed', i think that induction does not prove 'existence of a function, every term is not $1$'. – Katlus Oct 7 '12 at 22:02

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