Consider the power set of \mathbb{N}, that is, the set of all possible integer sequences a, where 1 \leq a_1 < a_2 < a_3 <... . If only infinite sequences are allowed, then these are homeomorphic to the interval (0,1] and therefore uncountable.

Proof: Define \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, P={2,3,5,7,11,13,17,19,...}, so \phi(P)=.0110101000101000101... \approx .4146825099 . Therefore, any subset of \mathbb{N} can be expressed uniquely by a real number \phi(a) \in [0,1] , \phi converges regardless of a, \phi(\mathbb{N})=.111111111...=1, and \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 \{1,4,6\} is the same as \{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 \mathbb{N} is uncountable, than so certainly is the power set of \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 \mathbb{N} is uncountable.

Is this way of interpreting infinite sequences known and/or applicable in math?