Let be $X$ and we assume that $(x_\lambda)_{\lambda\in\Lambda}$ is a net such that there exists a cofinal and increasing map $\varphi$ form $\Bbb N$ to $\Lambda$ such that $\big(x_{\varphi(n)}\big)_{n\in\Bbb N}$ converges to any $x_0\in X$: so if $X$ is first countable then the convergence of $\big(x_{\varphi(n)}\big)_{n\in\Bbb N}$ to $x_0$ implies the convergence of $(x_\lambda)_{\lambda\in\Lambda}$ to $x_0$? If the answer to the last question is negative then there is a relevant reason different form this or this or rather this that show why into first countable space is sufficent to consider only sequence?
If $X$ is first-countable then a net converges when a subsequence converges
examples-counterexamplesfirst-countablegeneral-topologynetssequences-and-series
Related Solutions
For first countable topological spaces, it is enough to consider convergence of sequences to determine closure of sets and continuity of functions. The following result can be found in several Topology books (Kelley's Genral Topology for instance. The section on Topology of "Hitchhiker's guide to Infinite Dimensional Analysis" by the late Aliprantis (a fantastic read) covers this in a very elegant way)
Theorem: If $(X,\tau)$ is first countable, then:
- $X$ is Hausdorff iff any convergent sequence in $X$ has a unique limit.
- A point $x\in X$ is a cluster point of a sequence $\{x_n:n\in\mathbb{Z}_+\}$ iff there exists a subsequence that converges to $x$.
- A sequence $x_n$ converges to $x$ iff every subsequence converges to $x$.
- $x\in\overline{A}$ iff there is a sequence $x_n\in A$ that converges to $x$.
- For any topological space $(Y,\tau')$ and function $f:X\rightarrow Y$, $f$ is continuous at $x$ iff for any sequence $x_n\xrightarrow{n\rightarrow\infty} x$, $f(x_n)\xrightarrow{n\rightarrow\infty} f(x)$.
- More generally, for any topological space $(Y,\tau')$ and function $f:X\rightarrow Y$, $\lim_{u\rightarrow x}f(x)=L$ iff $\lim_{n\rightarrow\infty}f(x_n)=L$ for any sequence $\{x_n:n\in\mathbb{N}\}\subset X$ such that $\lim_nx_n=x$.
Here is a sketht of the proof By hypothesis, any point $x\in X$ has a countable local base $\mathscr{V}_x=\{V_n:n\in\mathbb{N}\}$ and, by setting $U_n=\bigcap^n_{j=1} V_j$ if necessary, we may assume that $V_n\subset V_{n+1}$ for all $n\in\mathbb{N}$.
(1) Since any sequence is a net, only sufficiency remains to be proved. Suppose any convergent sequence in $X$ has a unique limit. Let $x$ and $y$ be points in $X$ and let $\{V_n:n\in\mathbb{N}\}$ and $\{U_n:n\in\mathbb{N}\}$ be decreasing local neighborhoods of $x$ and $y$ respectively. If $V_n\cap U_n\neq\emptyset$ for all $n\in\mathbb{N}$ then we can choose $x_n\in V_n\cap U_n$. The sequence $\{x_n:n\in\mathbb{N}\}$ converges to both $x$ and $y$. Therefore, $x=y$.
(2) Since a subsequence of a sequence is a subnet of the sequence, only necessity remains to be proved. Suppose $x$ is a cluster point of the sequence $\{x_n:n\in\mathbb{N}\}$. There is $n_1\geq 1$ such that $x_{n_1}\in V_1\in \mathscr{V}_x$. Having found $x_{n_1},\ldots, x_{n_k}$ such that $n_1<\ldots < n_k$ and $x_{n_j}\in V_j$ we choose $x_{n_{k+1}}\in V_{k+1}$ such that $n_{k+1}\geq n_k+1$, which is possible since $x$ is a cluster point of $\{x_n:n\in \mathbb{N}\}$. Therefore, $\{x_{n_k}:k\in\mathbb{N}\}$ is a subsequence that converges to $x$.
(3) This statement is trivial, try to complete it.
(4) Since any sequence is a net, only necessity remains to be proved. If $x\in \overline{A}$ then $V_n\cap A\neq\emptyset$ for each $V_n\in\mathscr{V}_x$. Choosing $x_n\in V_n\cap A$ for each $n\in\mathbb{N}$, we obtain a sequence $x_n\xrightarrow{n\rightarrow\infty} x$.
(5) Since any sequence is a net, only sufficiency remains to be proved. Suppose $f(x_n)\xrightarrow{n\rightarrow\infty} f(x)$ whenever $x_n$ is a sequence with $x_n\xrightarrow{n\rightarrow\infty} x$. If $f$ fails to be continuous at $x$, then there is a neighborhood $U\in\mathcal{V}_{f(x)}$ such that for any $n\in\mathbb{N}$ there is $x_n\in V_n$, $V_n\in\mathscr{V}_x$, with $f(x_n)\notin U$. Then $x_n$ is a sequence converging to $x$ for which $f(x_n)\nrightarrow f(x)$. This is a contradiction.
(6) By replacing $f(x)$ by $L$ in the proof of (5), the remaining of that roof carries over.
Best Answer
As Ruy suggested above into the comments the conjecture is generally false: e.g. the sequence $(x_n)_{n\in\Bbb N}$ defined as $$ x_n:=\begin{cases}0,\,\text{if }n\,\text{is even}\\n,\,\text{if }n\,\text{is even}\end{cases} $$ is a sequence into a first countable space that does not converege but it has trivially a converging subsequence.