let $s_{n}$ be an arbitrarily long sequence of binary numbers. Then $S$ is the set of all possible sequences with that length. It has a cardinality of 2^length. For example, sequences two digits long have a cardinality of 4.

Here is the proof by mathematical induction.

There exists a bijection between the first two natural numbers and the set of sequences with length 1. $s_{1}=0$ and $s_{2}=1$.

Assume there is a set of sequences that has a bijection to a subset of the natural numbers. Then the sequences are $s_{1}$ to $s_{(2^{length})}$. Then the set of sequences with length=$length+1$ has a bijection to a subset of the natural numbers. They are labeled $s_{1}$ to $s_{(2^{length+1})}$, indicating a bijection to the subset of the first $length+1$ natural numbers.

By mathematical induction, there exists a bijection between sequences of any length and a subset of the natural numbers. By analytical continuation, that, includes those of infinite length.