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Consider a convergent sequence of rational numbers $a_n$ with a limit $\lim_{n\rightarrow\infty} a_n = b$.

Does there exists some kind of necessary condition for $b$ to be rational that only uses the elements of the sequence?

It is easy to describe sufficient condition for integers, like $\forall n\in \mathbb N, \ a_n \in \mathbb Z \implies b \in \mathbb Z.$

But this is definitely not true for rationals: $\forall n\in \mathbb N, \ a_n \in \mathbb Q \not\implies b \in \mathbb Q.$

Also it is true that: $\forall n\in \mathbb N, \ a_n \in \mathbb R \implies b \in \mathbb R.$

Are there sufficient conditions for $b$ to be rational that are not of the form $$\exists k \in \mathbb N: |\{ n \in \mathbb N \ | \ a_n = a_k \}| = \infty?$$

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Yes, in other words $\mathbb{Z}$ and $\mathbb{R}$ are closed, but $\mathbb{Q}$ is not. I doubt there are such nontrivial conditions. – 1015 Feb 22 '13 at 3:33
@julien Yeah, but wouldn't it be awesome if there was? – Valtteri Feb 22 '13 at 19:45

It is not true that: $\forall n\in \mathbb N, \ a_n \in \mathbb R \implies b \in \mathbb R.$ (I assume that by $x\in \mathbb R$ you mean $x\in \mathbb R - \mathbb Q$)

For example $\ a_n = 1/n \pi$ converges to $0 \in \mathbb Q$ and $\ a_n = p/q + 1/n \pi$ converges to $p/q \in \mathbb Q$ for any p/q.

I am thinking that looking at the extended fraction representation might help in your initial question, as rational numbers the fraction terminates but for irrational numbers it is infinite.

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