Proving if limit exists then function is continuous The question I'm trying to solve:
Let $(X, d)$ and $(Y, e)$ be metric spaces, where $(Y, e)$ is complete.
Let $A \subseteq X$ be dense in $X$. Suppose $f: A \to Y$ is uniformly continuous.
Prove there exists a (unique) continuous map $F: X \to Y$ such that $F(x) = f(x)$ for all $x \in A$.
One of the parts defines $F: X \to Y$ by
$$
F(x) = \lim_{n \to \infty} f(a_n)
$$
for any sequence $(a_n) \in A$ which converges to $x$ in $(X, d)$.
The part that I'm struggling is:
Prove that $F: X \to Y$, as defined above, is continuous.
It clear that $F$ is continuous for all $x \in A$ as $F(x) = f(x)$ for $x \in A$ and it preserves uniform continuity on $A$ from $f$.
I'm trying to show continuity using definition in terms of limits of sequences which is as follows:
$$
\forall (a_n) \in A \lim_{x \to \infty} a_n = x \implies \lim_{x \to \infty} f(a_n) = f(x)
$$
I have shown before that sequence $(f(a_n))$ is Cauchy and therefore convergent in $(Y, e)$ as it's complete.
Therefore I know that the $\lim_{x \to \infty} f(a_n)$ exists and I denote it $L$.
All that is left to show is that $L = f(x)$ where $x \in A'$ ($A'$ denotes limit points of $A$).
I suspect this is true due to uniform continuity of $f$ which I suppose show us that $d(a_n, x) \implies e(f(a_n), f(x))$.
Yet, I'm not sure we could do this since f is defined on A and cannot take values of A'.
Thus I don't really understand how to show this.
 A: Let's prove this in stages. First, a simple fact (you stated this, but we'll prove it anyway):
Lemma. If the sequence $\{a_n\} \subset A$ converges to $x \in X$, then $\{f(a_n)\}$ is Cauchy in $(Y,e)$.
Proof. Indeed, since $f$ is uniformly continuous, for every $\varepsilon > 0$ we have $e(f(a),f(b)) < \varepsilon$ whenever $d(a,b) < \delta$. And since $\{a_n\}$ is Cauchy, there exists $N \geq 0$ such that $d(a_m, a_n) < \delta$ for all $m,n \geq N$. Thus, $e(f(a_m), f(a_n)) < \varepsilon$ for all $m,n \geq N$; this means that $\{f(a_n)\}$ is Cauchy. $\square$
Therefore, $\lim f(a_n)$ is a nice, well-defined object for sequences $a_n \to x$, since $(Y, e)$ is complete.
Next, let's show that $F$ is well-defined. That is, for $x \in X$ and sequences $\{a_n\}, \{b_n\} \subset A$  with $a_n \to x$ and $b_n \to x$, we show that
$$\lim f(a_n) = \lim f(b_n).$$
The important observation is that the sequence $\{c_n\} = \{a_1, b_1, a_2, b_2, \dots\}$, or formally by
$$
c_n = 
\begin{cases}
a_{(n+1)/2} & \text{if $n$ is even,} \\
b_{n/2} & \text{if $n$ is odd,}
\end{cases}
$$
also converges to $x$. The proof is to choose $N_1, N_2 \geq 0$ such that $d(x, a_n) < \varepsilon / 2$ when $n \geq N_1$ and $d(x, b_n) < \varepsilon / 2$ when $n \geq N_2$. Taking $N = \max(N_1,N_2)$ therefore suffices. It follows that $\{f(c_n)\}$ is Cauchy in $Y$ by our lemma, so $\lim f(c_n)$ exists, and it equals $\lim f(a_n)$ and $\lim f(b_n)$ by taking subsequences.
Now we turn to continuity of $F$. For $x_n \to x$ in $X$, we fix the sequences $\{a^n_k\} \subset A$ such that $a^n_k \to x_n$. Construct a new sequence $b_n$ as follows. Let $k_n \geq 0$ be the smallest integer such that $e(f(a^n_k), F(x_n)) < 1/n$ for all $k \geq k_n$. Set $b_n = a^n_{k_n}$ and observe that $\{b_n\} \subset A$. Then it follows from the triangle inequality that
$$
e(F(x_n), F(x)) \leq e(F(x_n), f(b_n)) + e(f(b_n), F(x)) < \tfrac{1}{n} + e(f(b_n), F(x))
$$
for each $n=1,2,\dots$. On the other hand, $b_n \to x$, so $e(f(b_n), F(x)) \to 0$.
Therefore $F(x_n) \to F(x)$ as $x_n \to x$.
The last part to show is uniqueness of $F$. Suppose $G$ is also a continuous extension of $f$ from $A$ to $X$. Let $x \in X$, and fix $\{a_n\} \subset A$ such that $a_n \to x$. Then
$$
F(x) = \lim F(a_n) = \lim f(a_n) = \lim G(a_n) = G(x),
$$
so in fact $F=G$.
