I am self studying Topology from Gemignani's Elementary Topology. Here's the question which I am trying to prove (Exercise 2 on page 127):

Let $X,D$ be a metric space and $\{ s_i \}, i \in I$ be a net in $X$. If every subsequence of $\{ s_i \}$ converges to $x$, then show that $\{ s_i \}$ converges to $x$.

Suppose that $\{ s_i \}$ does not converge to $x$. Now, we're trying to find a subsequence which does not converge to $x$. By the definition, there is a open set $U$ containing $x$ such that for all $i \in I$, $s_j \not\in U$ for some $j \in I$ with $i\le j$. With this, I can easily construct $k : \mathbb{N} \to I$ such that $k$ is monotone and $s_{k_n} \not\in U$ for all $n \in \mathbb{N}$. The only problem I am facing is to find $k$ which satisfies all the properties. I notice that I couldn't even use the "niceness" that metric spaces offer in construction of such a function $k$.

Can someone drop some hints so that I complete this problem? Thanks in advance.

  • 2
    $\begingroup$ Can you find the definition of subsequence in this text? It cannot mean subnet with domain $\Bbb N$ as we would logically expect, because then this (and your previous) question are false. $\endgroup$ – Henno Brandsma May 22 '20 at 14:39
  • $\begingroup$ If the net does not converge we have that there exists on open neighbourhood $U$ of $x$ such that $$\forall i \in I: \exists j: (j \ge i) \land (s_j \notin U)$$ But this gives you no chance to use special properties of metric spaces. $\endgroup$ – Henno Brandsma May 22 '20 at 14:43
  • $\begingroup$ @HennoBrandsma Author never formally defines a subsequence of a net. He "italicizes" the first time he makes use of it. drive.google.com/file/d/1821t1AD-V4mYAMiC5aYGQ4b0i8mHd4It/… . See exercise 4 (page 122), I've attached Example 1 as well. The book is partially available on Google Books. $\endgroup$ – ashK May 23 '20 at 5:15
  • $\begingroup$ If it helps, kindly let me know what the author is trying to mean by a "subsequence" of a net. $\endgroup$ – ashK May 23 '20 at 5:17
  • 1
    $\begingroup$ Well, consider the directed set of all functions from $\Bbb N$ to $\Bbb N$ ordered by almost dominance: $f \le g$ iff $\{n \in \Bbb N: g(n) > f(n)\}$ is at most finite. This is also a directed set with no countable cofinal set, so no subsequences. Ordinals and cardinals are not strictly needed. Also, a free ultrafilter on $\Bbb N$ doesn't have a countable cofinal subset under reverse inclusion-ordering etc. $\endgroup$ – Henno Brandsma May 24 '20 at 14:59

This is an a way a continuous of your previous question and Freakish's answer to it here is still relevant and shows that this statement is wrong: Let any $(x_{\alpha})_{\alpha \in \omega_1}$ be any net into $X$ defined on $\omega_1$, standard order. Then in Gemignani's definition, these net has no subsequences (convergent or otherwise) so vacuously we can say that "all subsequences of $(x_{\alpha})_{\alpha \in \omega_1}$ converge to $x$, whatever $x$ is. If your statement would hold we could conclude that $(x_{\alpha})_{\alpha \in \omega_1}$ converged to $x$, which would almost certainly be false (for all most all $x$ and spaces $(X,d)$).

In short, I'm sceptical. Maybe subsequence has a special meaning on this text (check the index to find the definition?), as @freakish suggested (so not a special subnet).


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