Fake induction proof Using the induction method:
$(\forall P)[[P(0) \land ( \forall k \in \mathbb{N}) (P(k) \Rightarrow P(k+1))] \Rightarrow ( \forall n \in \mathbb{N} ) [ P(n) ]]$
Why this proof is wrong?
$P(x)\equiv (\displaystyle\lim\limits_{a \to x}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q})$
Basis
$P(0)\equiv$ True ?
$P(0)\equiv(1\in\mathbb{Q})\equiv True$
Induction
$k \in \mathbb{N}$
$P(k)\equiv$ True $\implies P(k+1)\equiv$ True?
$P(k+1)\equiv$
$(\displaystyle\lim\limits_{a \to k}\sum_{i=0}^{a+1}\frac{1}{i!}\in\mathbb{Q}) \implies (\displaystyle\lim\limits_{a \to k}\sum_{i=0}^{a}\frac{1}{i!}+\lim\limits_{a \to k}\frac{1}{(a+1)!}\in\mathbb{Q})\implies (P(k)+\frac{1}{(k+1)!} \in \mathbb{Q})\implies$ True
So, $P(x)$ is true for all $x \in \mathbb{N}$.
But,
$\lim\limits_{x \to \infty}{P(x)}\equiv (\displaystyle\lim\limits_{x \to \infty}\lim\limits_{a \to x}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q})\equiv \lim\limits_{x \to \infty}\sum_{i=0}^{a}\frac{1}{i!}\in\mathbb{Q}\implies e \in \mathbb{Q}$. But we know $e$ don't belongs $\mathbb{Q}$.
So I would like to know, why this fake proof doesn't work, why run through all integers in $x$ until the last one, infinity, the formula fails.
Edit:
I had include limits to extend the discussion.
 A: No, your statement is true and the proof does work; it's just that rationality isn't preserved in the limit.  The key is that the statement is only for $n \in \mathbb{N}$, whereas $\infty \notin \mathbb{N}$ -- induction only proves things about natural numbers, each one of which is finite (even though there are an infinite number of them).
A: I would like to suggest a different perspective on the error in the argument - because it occurred to me, not because anything already said about induction and infinity is wrong.
If you construct the Real Numbers using Dedekind Sections you will see that every Real Number is the limit of a sequence of rational numbers. Real Numbers are useful in Analysis partly because they are closed under taking appropriate limits, while the Rational Numbers aren't. So one way of looking at the problem with the proof is that a limit has been taken in an inappropriate context.
A: You did prove $P(x)$ is true for all positive integers $x$. However that does not allow you to conclude that $P(\infty)$ is also true. Your example is a testament to this invalid conclusion, since it is known that $e$ is not rational.
