Cantor Set is Homeomorphic to the Countable Product of {0, 1}

Question

In Part of proof of homeomorphism from Cantor set ot infinite product of {0,1}

it is shown that the pre-image of sets of the form

$U(j,a)= \left\{ (a_n)_{n=1}^{\infty} ∈ \left\{ 0,1 \right\}^N :a_j=a \right\}$ where $j∈N$ and $a∈ \left\{0,1\right\}$

are open in C.

My question is, where does the $2.3^{-j}$ come from in the following part of the proof...

$\left\{x∈C:∀i≤j(x=h(i)\right\}=C∩[h^∗,h^∗+2.3^{−j})$ where $h^∗=\sum_{i=1}^j2h(i)3^{−i}$

Answer 1

Just to be clear, that should definitely be a multiplication sign and not a decimal point. As to why it is a multiplication sign, this follows from the definition of the Cantor Set, as we define it by iteratively splitting an interval $[a,a + 3^{-j+1}]$ into two intervals $[a, a + 3^{-j}]$ and $[a + 2\cdot 3^{-j}, a + 3^{-j + 1}]$. These are two copies of the same thing translated by that factor $2\cdot 3^{-j}$.