Inductive Reasoning

Considering that the truth of S_i implies the truth of S_i+1, i ε natural numbers, then starting from the truth of S₁, one can state the truth of S_n, n being any natural number. I was wondering whether we can make any statement as i→∞, such that the limS_i as i → ∞ is also true?

 

If we were to check the countability of the cartesian product between two countable sets, A = {a1, a2, a3, ....} and P = {p1, p2, p3, ....}, the resultant set should be countable: If were to consider
{(a1,p1), (a1,p2), (a1,p3), ....} = M₁
{(a2,p1), (a2,p2), (a2,p3), ....} = M₂
{(a3,p1), (a3,p2), (a3,p3), ....} = M₃
... ... ... ... ... ... ... ... ... ... ... ... ...
Taking the union of all the M_i's results in taking the union of a countable number of countable sets, and so the cartesian product A x P is countable. From this reasoning, we could say that for N = set of natural numbers, Nⁿ for some finite n is countable:
Nⁱ x N = Nⁱ⁺¹
N¹ x N = N² ---> since N is countable, so is N²
N² x N = N³ ---> since N² is countable, so is N³, and so forth.
However, if we were to continue this for Nⁿ as n→∞ (an infinite dimensional space), the countability is arguable (I initially thought that it would be countable by inductive reasoning):
Consider the set of real numbers [0,1). This set has power c of the real numbers. We can write each number as some decimal expansion 0.a1a2a3....
We can map these numbers into the infinite dimensional case of Nⁿ by making the correspondence 0.a1a2a3... → (a1,a2,a3,...). [0,1) would then be mapped to a proper subset of Nⁿ (since each decimal place is bounded between 0 and 9, and forms ending in 99999.... converge to some other form). Therefore Nⁿ can not be countable for the infinite dimensional case.
I had thought that induction implied that you could apply the condition on S_i onto limS_i as i → ∞, so I am unsure of whether there is a flaw in this proof somewhere.