# Power set of N

Consider the power set of $\displaystyle \mathbb{N}$, that is, the set of all possible integer sequences $\displaystyle a$, where $\displaystyle 1 \leq a_1 < a_2 < a_3 <...$ . If only infinite sequences are allowed, then these are homeomorphic to the interval $\displaystyle (0,1]$ and therefore uncountable.
Proof: Define $\displaystyle \phi(a) = \sum_{n=1}^{\infty} \frac{1}{2^{a_n}}$ . A visual way of looking at it is by expressing "in" and "out" of the set as bits in a binary string. Take the primes for example, $\displaystyle P={2,3,5,7,11,13,17,19,...}$, so $\displaystyle \phi(P)=.0110101000101000101... \approx .4146825099$ . Therefore, any subset of $\displaystyle \mathbb{N}$ can be expressed uniquely by a real number $\displaystyle \phi(a) \in [0,1]$ , $\displaystyle \phi$ converges regardless of a, $\displaystyle \phi(\mathbb{N})=.111111111...=1$, and $\displaystyle \phi(\null)=.0000000=0$ .
*Note: A terminating decimal .100101 can be expressed uniquely as a nonterminating decimal .100100111111111... by properties of decimal expansions. However, it is counterintuitive to think that the set $\displaystyle \{1,4,6\}$ is the same as $\displaystyle \{1,4,7,8,9,10,11,12,...\}$. But if we restrict ourselves to infinite subsets, we circumvent this problem. Besides, if the set of infinite subsets of $\displaystyle \mathbb{N}$ is uncountable, than so certainly is the power set of $\displaystyle \mathbb{N}$.
I am not sure how rigorous I've been, but is this a correct statement? The set of all possible ordered integer sequences bounded below by 1 is homeomorphic to the interval (0,1] and therefore the power set of $\displaystyle \mathbb{N}$ is uncountable.