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?
Best Answer
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:
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.