Consider the set $ X = \{(x_n): \sup n|x_n| < \infty\}$ with the metric $p(x,y) = \sum_{n=1}^\infty |x_n-y_n|/n$. Is the space $(X,p)$ complete? If not what is its completion?

In such questions I do not know how to start with or approach it. If it is not complete then I need to find a cauchy sequence that does not converge. If I take $x_n = 1/n$ then it is in the set X. But how can I say it it cauchy under this distance metric. I need to show $\exists N$ s.t. $\forall n,m>N$: $d(x_n,x_m)< \epsilon \ \forall\epsilon>0$ but I cannot figure out how to apply this $d(x_n,x_m)< \epsilon$ under the metric p.

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    $\begingroup$ Note that it is often simpler to find a convergent sequence (in an extension of $X$) but whose limit does not belong to $X$ because the limit does not verify the $\sup$ property. $\endgroup$ – zwim Dec 18 '17 at 1:17

The space is not complete. Let $\{x_n\}$ be any sequence such that $\{nx_n\}$ is not bounded but $\sum |x_n|^{2}<\infty$. For example, we can take $x_n=\frac {log {n}} n$. Consider the sequence $x^{(j)}$ in X where $x^{(j)}_n =x_n$ if $n \leq j$ and 0 otherwise. Coordinate wise this sequence in X converges to $\{x_n\}$ which does not belong to X. It is easy to check (using Cauchy-Schwartz inequality) that the sequence $x^{(j)}$ is Cauchy which does not converge in X.

  • $\begingroup$ The completion is $\{\{x_n\}:\sum \frac {|x_n|} n <\infty\}$ $\endgroup$ – Kavi Rama Murthy Dec 18 '17 at 8:23
  • $\begingroup$ Thank you for your answer but how did you find the completion? $\endgroup$ – Pumpkin Dec 18 '17 at 8:36
  • $\begingroup$ Finding the completion is guess work. To prove that the new space is indeed the completion you have to verify two facts: that the space is complete and the original space is dense in it. First of these is a routin e verification. For the second you just have to very that sequences with only finite number of non-zero elements are in the original space and they form a dense set in the new space. If you have trouble verifying these I can give some details but you should try it first. $\endgroup$ – Kavi Rama Murthy Dec 20 '17 at 10:09

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