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 $\{ s_i \} , i \in I$.
Yes. The same definition is given in the book.
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:
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.