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I only have some confusion with the following:

I)We say (X,d) is a complete metric space iff every Cauchy sequence in X converges to some x in X.

I know that every Cauchy sequence in R (real numbers) is convergent and by the above statement R is complete metric space. Now if we consider the open interval A=(-1,1), which is a subset of R, and the sequence a_n=(1+1/n), n>0 it converges to 1 and 1 not in A, and that would mean by I A is not complete metric space. So does that mean every subset of a complete metric space does not have to be complete metric space?

Second Could anyone please provide me another example of a Cauchy sequence in another metric spaces which is not convergent?

Thank you

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    $\begingroup$ I guess you probably mean $1 - 1/n$, not $1 + 1/n$, because otherwise it's not a sequence in $A$. You are correct that $(-1, 1)$ is not a complete metric space because of this issue. The standard example for your second question is $\mathbb{Q}$. $\endgroup$ – user296602 Nov 30 '17 at 0:18
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In order to proof that $A$ is not complete you have to choose a Cauchy sequence $a_n$ in $A$ (your $a_n$ do not lie in $A$, but if you choose $a_n=1-1/n$ this is a Cauchy sequence in $A$ that doesn't converge in $A$ so $A$ is not complete).

In general every closed subset of a complete space is again complete.

An example for an incomplete space is given by $\mathbb{Q}$ with the standard norm, because $\sqrt{2}\not\in\mathbb{Q}$ but there exists a Cauchy sequence in $\mathbb{Q}$ which converges to $\sqrt{2}$ and therefore it doesn't converge in $\mathbb{Q}$.

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