Why is there no bijection between the set of all sequences whose elements are the digits $0$ and $1$ and the positive integers? I have a problem understanding theorem 2.14 of Rudin's Principles of Mathematical Analysis. The theorem is the following:

$\textbf{Theorem 2.14:}$ Let $A$ be the set of all sequences whose elements are the digits $0$ and $1$. This set is uncountable. The elements of $A$ are sequences like $1,0,0,1,0,1,1,1,...$

$\textbf{Question:}$ Why there is not a $1$-$1$ mapping of the set $A$ to the set of all the positive integers?
I  would think that every positive integer (denoted by $J$ by Rudin) can be represented by an element of $A$ by using the binary representation. So each number is $$J_n = \sum_{i=0}^\infty s_{ni}2^i  $$
where $s_{ni}$ is the $i^\text{th}$ element of the sequence $s_n$ (using zero-based numbering).
Then, if $A\sim J$, shouldn't $A$ be countable?
Rudin mentions this theorem implies that the set of all real numbers is uncountable, so I understand this sequences actually represent real numbers and not the natural numbers, but why? I know my logic has at least one mistake (likely a huge blunder), but what is it?
 A: What would be the integer corresponding to the sequence $1,1,1,1,1,1...$ ?!
A: 
[these] sequences actually represent real numbers and not the natural numbers, but why?

Because they are infinitely long sequences, with the sequence $(s_{n,1}, s_{n,2}, s_{n,3}, \ldots)$ corresponding to the number
$$
\sum_{i=0}^\infty s_{ni}2^{-i}
$$
(note the minus sign in the exponent).
The set of all natural numbers can be placed in one-to-one correspondence with the set of all sequences of $0\text{s}$ and $1\text{s}$ in which there are only finitely many $1\text{s}.$ But the set that Rudin considers is not restricted to those containing only finitely many $1\text{s}.$
The non-existence of a one-to-one correspondence between this set and the set of all positive integers can be shown as follows. Suppose $s_{n,i}$ is the $i^\text{th}$ member of the $n^\text{th}$ sequence for $n\in \mathbb N.$ Then let
$$
t_i = 1 - s_{ii} \text{ for all } i \in \mathbb N.
$$
Then this sequence $t$ of $0\text{s}$ and $1\text{s}$ differs from every sequence corresponding as above to some natural number $n,$ because its $n^\text{th}$ digit differs from the $n^\text{th}$ digit of the $n^\text{th}$ sequence.
