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Proof that metric space of rational number with usual metric i.e $(\mathbb Q,d)$ is incomplete.

Attempt:- since we have a sequence of rational number $(1+1/n)^n$ converges to $e$, which is Cauchy but do not converge in $\mathbb Q$ (since $e$ is irrational ).

But how will I prove that this is Cauchy sequence! Can I use the fact that every convergent sequence is Cauchy? If yes, then the doubt is that the above sequence is convergent but not in $\mathbb Q$ and since when we say a Cauchy is not convergent we mean that it's is not convergent in the particular set ! So can we really use this fact of convergent implies Cauchy here !

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    $\begingroup$ The sequence is convergent in $\Bbb R$, so is definitely Cauchy. $\endgroup$ – Lord Shark the Unknown Nov 30 '19 at 19:35
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    $\begingroup$ You can use that the sequence is convergent to get that it is Cauchy in $(\mathbb{R}, d)$. Then it is also Cauchy in $(\mathbb{Q}, d')$ since the metric $d'$ is just the restriction of the standard metric on $\mathbb{R}$ to $\mathbb{Q}$. It may be worth formally proving this to yourself. $\endgroup$ – nbritten Nov 30 '19 at 19:36
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The fact that a sequence is a Cauchy sequence only depends upon the distances between its elements. So, if it is a Cauchy sequence in $\mathbb R$, it is also a Cauchy sequence in $\mathbb Q$. And it is a Cauchy sequence in $\mathbb R$ since it converges in $\mathbb R$.

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Let's say you have a sequence of rational numbers $a_n$ that is convergent in $\mathbb R$. Then it is a Cauchy sequence, meaning for any $\epsilon$ there exists $N$ such that $|a_m-a_n|<\epsilon$ for all $m, n>N$. These inequalities still hold in $\mathbb Q$, so while it may not converge in $\mathbb Q$, it is still Cauchy.

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