I am having trouble understanding a small part from theorem 3.2 ($1\implies 2$) of this paper (Between compactness and completeness, G. Beer).

Prerequisite: (1) A sequence $(x_n)$ in a metric space $(X,d)$ called cofinally Cauchy if for $\epsilon>0~\exists~\mathbb N_0\text{ (infinite)}\subset\mathbb N$ such that $d(x_i,x_j)<\epsilon~\forall~i,j\in\mathbb N_0.$

(2) A metric space $X$ is cofinally complete if every cofinally Cauchy sequence in $X$ clusters.

(3) Let $X$ be a metric space and $x\in X.$ If $x$ has a compact neighbourhood set, $v(x)=\sup\{\epsilon>0:S_\epsilon[x]\text{ is compact}\},$ otherwise set $v(x)=0.$

Main theorem 3.2: (1) X is cofinally complete $\implies$ (2) Whenever $(x_n)$ is a sequence in $X$ with $\lim_{n\to\infty}v(x_n) = 0,$ then $(x_n)$ has a cluster point:

Suppose condition (2) fails, and let $(x_n)$ be a sequence in $X$ with $\lim_{n\to\infty}v(x_n) = 0$ that has no cluster point. Without loss of generality, we can assume by passing to a subsequence that $ν(x_n) < 1/n$ and so we can find a sequence $(w^n_j)$ in $S_{1/n}[x_n]$ with no cluster point. Partition $\mathbb N$ into an countable family of infinite subsets $\{K_n: n\in\mathbb N\}.$ Then the assignment $y_j = w^j_n$ for $j\in K_n$ defines a cofinally Cauchy sequence $(y_j)$ in $X$ without a cluster point. Thus (1) fails.

I have verified the theorem other than the fact that $(y_j)$ is cofinally Cauchy. Please tell me why $(y_j)$ is cofinally Cauchy?

Thank you in advance.

  • $\begingroup$ MathOverflow copy of the question: Why $(y_j)$ is cofinally Cauchy? $\endgroup$ – Martin Sleziak Jun 16 '18 at 5:18
  • $\begingroup$ I should say that you have made (in my opinion) a good job in making the question readable without needing to read the linked paper. So +1 from me. (Maybe mentioning the suspected typo - which you already mentioned in comments under my answer - could be additional improvement of the question.) $\endgroup$ – Martin Sleziak Jun 16 '18 at 12:57
  • $\begingroup$ Thank you for your support and cooperation $\endgroup$ – Jave Jun 16 '18 at 12:59
  • $\begingroup$ I can only guess why the somebody decided to downvote the question - since the downvoter did not leave any comment. One thing I can think of is that you only mentioned the possible typo in comments and this information is, somewhat relevant. Another possibility is that the downvoter simply does not like cross-posting. $\endgroup$ – Martin Sleziak Jun 16 '18 at 13:03
  • $\begingroup$ For your own benefit you can have a look at discussions about cross-posting on meta. This answer offers some good advice. In a discussion on Meta MathOverflow it was suggested to wait at least a week $\endgroup$ – Martin Sleziak Jun 16 '18 at 13:03

EDIT: This is the original version of the answer, which was posted based on the original revision of the question where we had $y_j=w^j_n$. See also some comments below this answer.

From the construction of the sequence $(y_j)$ you can see, that if $j\in K_n$ then you have $y_j\in S_{1/n}[x_n]$. So you have $$d(y_i,y_j)<\frac 2n$$ for any $i,j\in K_n$. Since each $K_n$ is infinite (and thus cofinal in $\mathbb N$), you get that $y_j$ is cofinally Cauchy.

  • $\begingroup$ In the paper they have taken the sequence to be $(w^n_i)$ not $(w^i_n).$ Is it a printing mistake? $\endgroup$ – Jave Jun 16 '18 at 5:23
  • $\begingroup$ @Jave Actually, the way it is written in the current revision of your question and also in the paper it makes sense, as far as I can tell. I did not notice the discrepancy when posting the answer, sorry for that. $\endgroup$ – Martin Sleziak Jun 16 '18 at 5:38
  • $\begingroup$ How does it make sense under the current revision? $\endgroup$ – Jave Jun 16 '18 at 5:40
  • $\begingroup$ I mean how can we show it cofinally Cauchy under current revision $\endgroup$ – Jave Jun 16 '18 at 5:41
  • 1
    $\begingroup$ Sorry @Jave, I am not able to resolve this now. (But I agree with you that it seems that this way there might be a problem.) I have some other stuff to attend to IRL. If I have time later, I'll try to get back to this. (Well, unless somebody else answers you in the meantime.) $\endgroup$ – Martin Sleziak Jun 16 '18 at 5:53

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