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It is not hard to find proofs showing that $\mathbb{Q}$ is not complete with respect to the metric induced by the valuation $|\;\;|_p$. For example, it is enough to recall that every complete metric space is Baire and to show that $(\mathbb{Q},|\;\;|_p)$ is not Baire. Another option is to show that $\mathbb{Q}_p$ is uncountable, so $(\mathbb{Q},|\;\;|_p)$ is different than its completion. Both arguments use the fact that $\mathbb{Q}$ is countable.

But when I try to justify that $k(x)$ is not complete in the metric induced by the $x$-adic valuation $|\;\;|_x$, I don't find ideas or references addressing the proof. How can I proof the incompleteness of $(k(x), |\;\;|_x)$ without invoking its completion field $k((x))$? Any reference?

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  • $\begingroup$ What is the limit of the Cauchy sequence $3$, $3+x$, $3+x+4x^2$, $3+x+4x^2+x^3$, $3+x+4x^2+x^3+5x^4$... where the coefficient of $x^n$ is the $n$th decimal digit of $\pi$? $\endgroup$
    – Anthony Quas
    Commented Nov 19, 2016 at 1:38
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    $\begingroup$ I don't know. Are you suggesting that if we consider the elements of $k(x)$ as Laurent power series in $k((x))$, then the coefficients should satisfy a distinctive property like periodicity? If yes, can you please mention any reference to read? $\endgroup$
    – Chilote
    Commented Nov 19, 2016 at 1:43
  • $\begingroup$ I haven't worked through the details but if k has characterstic $0$, then $\exp(x)$ can't be rational by looking at derivatives after killing it's denominator. To push this through to a complete proof, you would have to show that if a series converges to a rational function, then the derivatives of the terms converges to the derivative of the rational function. I don't know what to do about positive characteristic. $\endgroup$
    – Asvin
    Commented Nov 19, 2016 at 4:02
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    $\begingroup$ Keyword: linear recurrence relation. See also math.stackexchange.com/questions/1720517/…, math.stackexchange.com/questions/228422/… $\endgroup$
    – YCor
    Commented Nov 19, 2016 at 4:20
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    $\begingroup$ I have a written proof in math.stackexchange.com/q/2025736 I couldn't do it without invoking $K((x))$. $\endgroup$
    – Chilote
    Commented Nov 22, 2016 at 13:00

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