I'm self studying Topology from Michael Gemignani's Elementary Topology. The author asks the following question (Exercise 2 on page 127):

Suppose $X,D$ is a metric space and $\{ s_i \} , i \in I$, is a net in $X$ and suppose that $s_i \to x$. Prove that a subsequence of $\{ s_i \} , i \in I$, converges to $x$.

I'm not sure what the author means by the subsequence of a net. The author defines subnets. Is subsequence of a net is a subnet indexed by the directed set $\mathbb{N}$? If so, we're done since we know that every subnet will converge to the same point where the net does.

Here's Gemignani's definition of subnet:

Let $\{ s_i \} , i \in I$ be a net in $X$. Let $J$ be a directed set and $k: J \to I$ such that

  • if $j \le j'$ then $k(j) \le k(j')$

  • if $i, i' \in I$, then there is $j \in J$ such that $i \le k(j)$ and $i' \le k(j)$.

The composition $s \circ k$ is said to be subnet of the net $\{ s_i \} , i \in I$.

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    $\begingroup$ I'm not sure if this is what the author means, but recall that sequences are particular cases of nets. Thus, a subsequence of a net could mean 'a subnet of the net which is actually a sequence'. $\endgroup$ – Manuel Norman May 22 at 9:38
  • $\begingroup$ What is Gemignani's definition of a subnet? $\endgroup$ – Henno Brandsma May 22 at 11:21
  • $\begingroup$ @HennoBrandsma I've updated in the description of my question. $\endgroup$ – Ashish K May 22 at 11:27

Is subsequence of a net is a subnet indexed by the directed set $\mathbb{N}$?

Yes. The same definition is given in the book.

If so, we're done since we know that every subnet will converge to the same point where the net does.

That depends. If the author literally meant "any subsequence of $(s_i)$ converges to $x$" then this is trivially true, as you've noted. Note that if $(s_i)$ has no subsequences (which can happen) then this is vacuously true.

However this interpretation doesn't bring anything new to the table. It seems that the author meant "$(s_i)$ has a subsequence convergent to $x$". Which unfortunately is not true in general. Recall the definition of a subnet:

Definition: For a net $f:I\to X$ a subnet of $f$ is a net $g:J\to X$ together with a monotone, cofinal function $h:J\to I$ such that $f\circ h=g$.

Cofinal here means that the image of $h$ is cofinal with $I$.

And thus the second interpretation of the statement is false. Simply because there are ordinals of uncountable cofinality. I.e. there is an ordinal $\lambda$ such that no countable subset of $\lambda$ is cofinal with $\lambda$. And so a constant net $t:\lambda\to X$, $t(i)=x$ is obviously convergent, but it has no countable subnet. Note that a constant sequence $x_n=x$ is not a subnet of $t$ - the cofinality axiom is not satisfied.

I think what the author wanted to say is that in metric spaces nets can be replaced with sequences. Meaning if $s:I\to X$ is a convergent net then there is a sequence $t:\mathbb{N}\to X$ convergent to the same limit and such that $im(t)\subseteq im(s)$. But such sequence need not be a subnet. Being a subnet is a stronger condition. For metric spaces pretty much any property requiring a net, can be stated with sequences instead.

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  • $\begingroup$ The question is unclear:"prove that a subsequence of $s_i$ converges to $x$ " is true as stated: it actually means "if we have a subsequence, then it converges to $x$" which is indeed obvious. But the author of the text certainly meant that such a subsequence exists, which is a novel element, and true of course. $\endgroup$ – Henno Brandsma May 22 at 11:07
  • $\begingroup$ @HennoBrandsma Indeed, this is unclear. I've updated the answer. I don't think a subsequence needs to exist. $\endgroup$ – freakish May 22 at 11:12
  • $\begingroup$ @freakish I think the statement holds (for any first-countable $T_1$ space) once you add the hypothesis that the net in question is not eventually constant. So while your counterexample is technically correct, it is essentially the only counterexample. $\endgroup$ – tomasz May 28 at 8:51

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