Why doesn't diagonalization prove that integers are not countable? I understand how Cantor's diagonalization argument works with respect to disproving that a bijection between integers and real numbers can exist. What I don't get is why the same reasoning doesn't apply to a bijection from integers to integers. Consider the argument as described here:
http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument#An_uncountable_set
My intuition is telling me that the same thing is not possible because the representation of a given integer is finite in length. I realize that this question has been asked before here: Why Doesn't Cantor's Diagonal Argument Also Apply to Natural Numbers?
But I found the answers somewhat opaque. That's why I am asking again.
 A: As it has been explained in those answers, Cantor's diagonal argument works because we're working with strings of $0$s and $1$s under no constraint at all, whereas natural numbers will eventually "zero out", that is, the sequence will eventually become an infinite string of zeroes. Thus, if you're working with all of $$\{(a_0,a_1,\ldots):a_i\in\{0,1\})=2^{\Bbb N}$$ then the counterexample produced by Cantor's algorithm will again be an element of the sets of sequences of zeroes and ones, but if you're working with a proper subset of this, namely  $$\{(a_0,a_1,\ldots):a_i\in\{0,1\} \text{ and } a_n=0 \text{ eventually} )\subsetneq 2^{\Bbb N}$$
it might be the case the "counterexample" is not one at all, since it will not be eventually zero. Thus the argument fails. It is not too hard to come up with such problem:
Consider $2,4,8,16,\ldots$. This is $(0,1,0,\ldots)$, $(0,0,1,0,\ldots)$, $(0,0,0,1,\ldots),\ldots$ and Cantor dictates we take $(1,1,1,1,\ldots)$ as a counterexample. But $(1,1,1,1,\ldots)$ is certainly not eventually zero!
