# Question on metric spaces and nets

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.

• 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. – Henno Brandsma May 22 '20 at 14:39
• 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. – Henno Brandsma May 22 '20 at 14:43
• @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. – ashK May 23 '20 at 5:15
• If it helps, kindly let me know what the author is trying to mean by a "subsequence" of a net. – ashK May 23 '20 at 5:17
• 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. – 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 $$(X,d)$$).