I am reading the proof for the following theorem:

The sets$\displaystyle P(\mathbb{N})$and$\displaystyle \mathbb{R}$are numerically equivalent.

I understood the part where the author proved the existence of the one-to-one function $\displaystyle f: (0,1) \rightarrow P(\mathbb{N})$, but I find it hard to understand the converse; i.e., the proof of the existence of the one-to-one function $\displaystyle g: P(\mathbb{N})\rightarrow (0,1)$.

It says as follows :

and it continues to show that $\displaystyle g$ is one-to-one.We define a function $\displaystyle g: P(\mathbb{N}) \rightarrow (0,1)$. For $\displaystyle S\subseteq \mathbb{N}$, define $\displaystyle g(S)=0.s_1s_2s_3\cdot\cdot\cdot$, where

$\displaystyle S_n=\left\{ \begin{array}{rcl}

1 & \mbox{if}&n\in S\\

2&\mbox{if} & n \notin S

\end{array}\right.$

Thus $\displaystyle g(S)$ is a real number in (0,1), whose decimal expansion consists only of 1s and 2s.

Question: Let's say $\displaystyle S=\{1,2,3,...,n\}$. What do the first two numbers look like? Please show examples.