Context : I consider $(E, d_E)$ and $(F, d_f)$ two metric spaces and $A\subset E$ a dense subset of $E$ (i.e $\bar{A}=E$). The function $f$ is defined only on $A$.
I would like to prove that if I have two sequences $x_n$ and $v_m$ in $A$ that converges to the same point $x$ (not in A but in E), then the sequences $f(x_n)$ and $f(v_m)$ have the same limit provided that $f$ is uniformly continuous and that the space of arrival $(F, d_F)$ is complete.
Here is my attempt :
Consider $(x_n)$ and $(v_m)$ two sequences converging to the same point $x$. We have that
- $\exists N_1 : n>N_1\implies d(x_n, x)\leq \delta/2$
- $\exists N_2 : m>N_2\implies d(v_m, x)\leq \delta/2$
Thus for all $n,m>max(N_1, N_2)$ we have
$$ d(x_n, v_m)\leq d(x_n, x) + d(v_m, x)\leq\delta $$
Using the uniform continuity of $f$, we know that for $\varepsilon/3$ there exists a $\delta$ such that for all $n,m> max(N_1, N_2)$
$$ d(x_n, v_m)\leq\delta \implies d(f(x_n), f(v_m))\leq\varepsilon/3 $$
Now, we denote $y$ and $y*$ respectively the limit of $f(x_n)$ and $f(v_m)$ (the limit is well defined since the space of arrival is complete) we remark that :
- $\exists N_3 : n>N_3\implies d(f(x_n), y)\leq\varepsilon/3$
- $\exists N_4 : m>N_4\implies d(f(v_m), y*)\leq\varepsilon/3$
Thus, for all $\varepsilon>0$, we can find a $N=max(N_1, N_2, N_3, N_4)$ such that for all $n,m>N$ we have
$$ d(y,y*)\leq d(y, f(x_n)) + d(f(x_n),f(v_m)) + d(f(v_m), y*)\leq \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon $$
wich shows that $d(y,y*) = 0$ so $y=y*$.
I would like to know if the proof seems correct, my intuition was to make a link between the two sequences using the uniform continuity and I tried to make this argument valid so please don't hesitate to correct me or add your proof !
EDIT
I missed some important information about the context I am very sorry.
This new problem is equivalent to say that you can extend the function $f$ to a new continuous function $tilde f: E \rightarrow F$. Indeed if $f:A \rightarrow F$, one might choose to extend it by $\tilde f: E \rightarrow F$, setting $\displaystyle \tilde f(x):= \lim_{\substack{a \rightarrow x \\ a \in A}} f(a)$. But the well-definedness of $\tilde f$ requires to show if $a_n \rightarrow x$, $b_n \rightarrow x$, and $a_n, b_n \in A$, then $\lim f(x_n) = \lim f(v_n)$ which is exactly your problem.
Note that $A$ being dense is important in the definition since you should be able to converge to $x \in E$ by a sequence $a_n \in A$ in the first place.
Take $y_n:= f(a_n)$ and $z_n:= f(b_n)$. As $F$ is complete, cauchy sequences will converge. So let us show $y_n$ and $z_n$ are cauchy sequences. Let $\epsilon > 0$. By $\color{red}{\text{uniform continuity}}$, there is some $\delta > 0$ so that if $d(p, q) < \delta$, $d(f(p), f(q)) < \epsilon$. As $a_n \rightarrow x$, there is $N_1 \in \mathbb{N}$ so that for all $m, n \ge n$, $d(a_{\color{red}{m}}, a_{\color{red}{n}}) < \delta$. (Note that convergent sequences are Cauchy sequences.) Therefore $d(f(a_m), f(a_n)) < \epsilon$.
This shows $y_n$ is Cauchy. Similarly you can prove $z_n$ is Cauchy. Therefore there is $y \in F$ so that $y_n \rightarrow y$ and $z \in F$ so that $z_n \rightarrow z$.
Finally, lets show that $y = z$. This time we use the fact that $a_n$ and $b_n$ are both converging to the same point $x$. Therefore there is $M \in \mathbb{N}$ so that for all $n \ge M$, $a_n, b_n \in B_{\delta/2}(x)$ and therefore $d(a_n, b_n) < \delta$. Again, by uniform continuity, $d(f(a_n), f(b_n)) < \epsilon$ for all $n \ge M$.
To summarize, $d(y_n , z_n) < \epsilon$ for all $n \ge M$. Apply the triangle inequality to this, the fact that $y_n \rightarrow y$ and $z_n\rightarrow z$ to conclude that $d(y, z) < \epsilon$. As $\epsilon$ is arbitrary, you get the result.