If $X$ and $Y$ are first countable then $\lim_{x\to x_0}f(x)=y_0$ iff $\lim_{n\to+\infty}f(x_n)=y_0$ for any $x_n\to x_0$ Definition
The limit of a function $f:X\to Y$ as $x$ approaches at the limit point $x_0$ is $y_0$ if and only if any net $\nu:\Lambda\to X\setminus\{x_0\}$ converging to $x_0$ is such that $f\circ\nu$ converges to $y_0$.
Lemma
If $\nu:\Lambda\to X$ is a net with values in a hausdorff space then it can converges at most one point.
Theorem
If $f:X\to Y$ is a function beetween hausdorff spaces then the limit of $f$ as $x$ approaches at $x_0$ is unique.
I know (is it true?) that if $f:\Bbb R^n\to\Bbb R^m$ is a function then the limit of $f$ as $x$ approaches at $x_0$ is $y_0$ if and only if for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $f(x_n)$ converges to $y_0$.
So clearly by definition I gave above if the limit of $f$ as $x$ approaches at $x_0$ then for any sequence (a sequence is a net!) $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it must be that $f(x_n)$ converges to $y_0$. Unfortunately I can't prove the inverse but I have had the following idea.
Conjecture
If $X$ and $Y$ are first countable and if $f:X\rightarrow Y$ is a function then the limit of $f$ as $x$ approaches at $x_0$ is $y_0$ if and only if for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $f(x_n)$ converges to $y_0$
So is the conjecture true? If not then if I add some hypotheses (hausdorff separability?) then is it true? So could seomeone help me, please?
 A: Theorem
If $X$ is first countable and if $f:X\to Y$ is a function then $y_0$ is the limit of $f$ as $x$ approaches at $x_0$ if and only if for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $\left(f(x_n)\right)_{n\in\Bbb N}$ converges to $y_0$.
Proof. If $y_0$ is the limit of $f$ as $x$ approaches at $x_0$ then by definition of limit for any net $(x_\lambda)_{\lambda\in\Lambda}$ converging to $x_0$ it happens that $\left(f(x_\lambda)\right)_{\lambda\in\Lambda}$ converges to $y_0$ so that clearly for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it must be $(f(x_n))_{n\in\Bbb N}$ converges to $y_0$ because any sequence is a net.
Conversely now we suppose for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $\left(f(x_n)\right)_{n\in\Bbb N}$ converges to $y_0$. So we proceed with reductio ad absurdum and we suppose that there exist a net $(x_\lambda)_{\lambda\in\Lambda}$ converging to $x_0$ whose image $\left(f(x_\lambda)\right)_{\lambda\in\Lambda}$ does not converges to $y_0$ so that there exist a neighborhood $U_{y_0}$ of $y_0$ such that for any $\lambda\in\Lambda$ there exist $\overline{\lambda}\ge\lambda$ such that $x_{\overline{\lambda}}\notin U_{y_0}$ and so we conclude that $\left(f(x_\lambda)\right)_{\lambda\in\Lambda}$ is frequently in $Y\setminus U_{y_0}$. Now we define
$$
\overline{\Lambda}:=\{\overline{\lambda}\in\Lambda:\overline{\lambda}\ge\lambda\,\,\,\wedge\,\,\,x_{\overline{\lambda}}\in Y\setminus U_{y_0}\}
$$
and we prove that it is a directed set with the induced relation. So clearly the induced relation is reflexive and transitive and then if $\,\overline{\lambda}_1,\overline{\lambda}_2\in\overline{\Lambda}$ so we consider $\lambda_3\in\Lambda$ (remember that $\Lambda$ is a directed set) such that $\lambda_3\ge\overline{\lambda}_1,\overline{\lambda}_2$ and so $\overline{\lambda}_3\in\Lambda$ (remember that $\left(f(x_\lambda)\right)_{\lambda\in\Lambda}$ is frequentely in $Y\setminus U_{y_0}$) such that $\overline{\lambda}_3\ge\overline{\lambda}_1,\overline{\lambda}_2$ so that we conclude that $\overline{\Lambda}$ is a directeed set. Now we define a function $\iota:\overline{\Lambda}\rightarrow\Lambda$ through the condiction
$$
\iota(\overline{\lambda}):=\overline{\lambda}
$$
for any $\overline{\lambda}\in\overline{\Lambda}$ and we prove that it is increasing and cofinal so that $\left(x_{\overline{\lambda}}\right)_{\overline{\lambda}\in\overline{\Lambda}}$ will be a subset of $(x_\lambda)_{\lambda\in\Lambda}$. So clearly if $\overline{\lambda}_1,\overline{\lambda}_2\in\overline{\Lambda}$ are such that $\overline{\lambda}_1\le\overline{\lambda}_2$ then $\iota(\overline{\lambda}_1)=\overline{\lambda}_1\le\overline{\lambda}_2=\iota(\overline{\lambda}_2)$ and so we remember that $\left(f(x_\lambda)\right)_{\lambda\in\Lambda}$ is frequentely in $Y\setminus U_{y_0}$ so that for any $\lambda\in\Lambda$  there exist $\overline{\lambda}\in\Lambda$ such that $\overline{\lambda}\ge\lambda$ and $f(x_{\overline{\lambda}})\in Y\setminus U_{y_0}$ and so clearly $\iota$ is cofinal. So we have found a subnet $\left(x_{\overline{\lambda}}\right)_{\overline{\lambda}\in\overline{\Lambda}}$ converging naturally to $x_0$ (indeed $(x_\lambda)_{\lambda\in\Lambda}$ converges to $x_0$) whose image does not converges to $y_0$. Now $X$ is first countable and so we can consider a local countable base $\mathscr B(x_0):=\{B_n\in\mathcal U(x_0):n\in\Bbb N\}$ of $x_0$ so that if $\left(x_{\overline{\lambda}}\right)_{\overline{\lambda}\in\overline{\Lambda}}$ converges to $x_0$ then for any $n\in\Bbb N$ there exist $\overline{\lambda}_n\in\overline{\Lambda}$ such that $\overline{\lambda}\in B_n$ for any $\overline{\lambda}\ge\overline{\lambda}_n$ and so for any $n\in\Bbb N$ we define
$$
x_n:=x_{\overline{\lambda}_n}
$$
determining another subnet of $(x_\lambda)_{\lambda\in\Lambda}$ that in particular is a succession naturally converging to $x_0$. So finally we observe that $f(x_n)\notin U_{y_0}$ for any $n\in\Bbb N$ so that the image $\left(f(x_n)\right)_{n\in\Bbb N}$ of the sequence $(x_n)_{n\in\Bbb N}$ does not converge to $y_0$ and this is impossible for the hypotesis we have gave above so that the lemma holds.
A: 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.
