When do we need to show a function is well-defined?

Question

Proposition: Let $A$ be a dense subset of $B$. Let $f$ be a uniformly continuous function from $A$ to a complete metric space $C$. Then $f$ has a unique extension to a continuous function from $B$ to $C$.

My attempt, only on showing $f$ has an extension to a continuous function from $B$ to $C$:

Since $(A,d)\subseteq (B,d)$ such that $\overline{A}=B$, then $\forall b\in B$, given any $\epsilon \gt 0$, there exists $a\in A$ such that $a \in S_{\epsilon}(b)$. Due to Theorem 36, there exists a sequence $\{\alpha_n\}$ that converges to $b$. By Theorem 43, $\{\alpha_n\}$ is a Cauchy sequence. By Theorem 51, since $f$ is continuous, $\{f(a_n)\}$ is also a Cauchy sequence in $C$. Since $C$ is complete, $\{f(\alpha_n)\}$ converges to some $\gamma \in C$; Hence, $f(b)=c$.

But, I was told that this attempt at a proof is insufficient to prove the claim since I still need to show that $f$ is well-defined; which is also what the proof from Kaplansky's text has done.

Question 1: I do not understand why proving $f$ is well-defined is necessary due to my understanding that since $\{a_n\}$ arbitrary in the set $S$ of sequences that converges to $b$, then we no longer need to consider another sequence in $S$. But apparently, I am wrong. Why?

Question 2: Why is it allowed to assumed that the sequence at (16), seen below has a subsequence converging to $c$?

Answer 1

OK, here is my comment as an answer:

When you set $f(b)=\lim f(a_n)$, you have chosen the limit given by that subsequence. Your friend might have chosen a different sequence $(\alpha_n)$ with $\alpha_n\to b$ and would like to set $f(b)=\lim f(\alpha_n)$. You both have the same definition only if $\lim f(a_n)=\lim f(\alpha_n)$, i.e. if your definition does not depend on the arbitrary choice of a sequence converging to $b$ that you made. You know that for any arbitrary $a_n\to a$,$(f(a_n))$ converges; you do not yet know that all of the sequences $(f(a_n))$ converge to the same limit.

Answer 2

Q1: If $(a_n)$ and $( \alpha_n)$ are sequences whic both converge to $b$, then we know that the sequences $(f(a_n))$ and $f(\alpha_n))$ are convergent. In order to show that $f(b)$ is well-defined, we have to show that

$$\lim f(a_n)= \lim f(\alpha_n).$$

Q2: $(f(a_n))$ is a subsequence of the sequence in $(16)$ and $\lim f(a_n)=c$.

Answer 3

I don't see why it should be necessary to show that our constructed extension is well-defined, at least not because of different choices of sequences.

Define one extension $\bar f : A \to C$ by the following procedure: For every $a \in A$ set $\bar f(a) = f(a).$ For every $b \in B \setminus A$ take one sequence $(a_k) \subset A$ converging to $b$ (such a sequence exists by denseness of $A$) and set $\bar f(b) = \lim f(a_k).$ Here we need to show that the limit exists (this you might call well-definedness), but right now we don't need to show that it does not depend on the choice of sequence.

Now we have an extension. We have two steps left: 1) to show that $\bar f$ is continuous, 2) to show that it is unique. One way to show the latter might be to show that it doesn't depend on the choices of sequences.

Am I wrong? Do I miss something?