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

  • 1
    $\begingroup$ 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

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 $\mathbb{Q}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.