# Inductive Reasoning

1. Nov 15, 2011

### Gear300

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

2. Nov 15, 2011

### Stephen Tashi

I'm tempted to say "No" immediately because one interpretation of what you mean is "if a statement is true for any given natural number then it is true for the entire set of natural numbers". That wouldn't be valid reasoning. For example: S_i = "There is a natural number greater than i" versus "There is a natural number greater than any number in the set of natural numbers".

What you said mentions: " $lim_{n \rightarrow \infty} S_i"$ and you didn't define what this means. If we take "1" as representing "true" and interpret the limit as a the limit of a numerical sequence, then the limit is 1. But that is a different interpretation than in the previous paragraph.

3. Nov 15, 2011

### Gear300

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 Mi'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 Si onto limSi as i → ∞, so I am unsure of whether there is a flaw in this proof somewhere.

Last edited: Nov 15, 2011
4. Nov 15, 2011

### Stephen Tashi

That phrase has an intuitive appeal, but to reason precisely about it you have to give it a precise definition. What does it mean? After all, we can say that the limit of a numerical sequence is the result of "continuing a process to infinity", but you can't reliably use that definition in proofs. You have to use the epsilon-delta definition.