Incomplete metric space .

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 !

• The sequence is convergent in $\Bbb R$, so is definitely Cauchy. – Lord Shark the Unknown Nov 30 '19 at 19:35
• 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. – nbritten Nov 30 '19 at 19:36

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 $$\mathbb R$$.
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.