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In the metric space $\mathbb{Q}$ with the absolute value metric, let $x_n$ be the rational number $1.i_1 \ldots i_n$ which agrees with the decimal expansion of $\sqrt{2}$ to $n$ decimal places. Does the sequence $\{x\}_{n=1}^{\infty}$ converge?

So far I have is that the set $\{x\}_{n=1}^{\infty}$ is a monotone sequence which is bounded below by $1.4$ and bounded above by $\sqrt{2}$ and since $\{x_n\}$ is a bounded and monotone sequence therefore by the monotone convergence theorem then consequently then the sequence does converge to $\sqrt{2}$ . QED

It seems that the prove is too simple and that nothing was done with the metric space $\mathbb{Q}$ and the absolute value metric and some representation of $1.i_1 \ldots i_n$ in the proof.

Tell me if there is something that I should do to incorporate it into the prove or if my proof is completely wrong.

Thanks!

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    $\begingroup$ The question is whether the sequence converges in $\Bbb Q$. $\endgroup$ Sep 8 '15 at 21:25
  • $\begingroup$ The absolute value metric is just the usual metric. Perhaps you should still explain why $\sqrt2$ is the limit: so far you only stated that $\sqrt2$ is an upper bound. $\endgroup$
    – Berci
    Sep 8 '15 at 23:45

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