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http://www.scribd.com/mobile/doc/76236535

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 $ran$.2012-10-07

1 Answers 1

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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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    @Ayman I still don't get how can this be extended to $\mathbb{N}$.. I'll post it as another question. Thanks for helping me.2012-10-07