Uniform convergence of a sequences of functions to a complete metric space

I would appreciate clarification regarding Theorem 2.5 (page 10) in Gamelin and Greene "Intro to Topology."

The theorem states:

Let $S$ be a set and let $X$ be a complete metric space. If $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequences from $S$ to $X$, then there exists a function $f$ from $S$ to $X$ such that $\{f_n\}$ converges uniformly to $f$.

The proof goes as follows:

For each fixed $s \in S,$ $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequence in $X$. Since $X$ is complete, $\{f_n\}$ converges to some point of $X$, which we define to be $f(s)$.

Let $\epsilon>0$. Choose an integer $N$ such that $d(f_n(s),f_m(s))< \epsilon$ for all $s \in S$ and all $n,m \geq N$. Then

$$d(f_n(s),f(s)) \leq d(f_n(s),f_m(s)) + d(f_m(s),f(s))$$

and $$d(f_n(s),f(s)) < \epsilon + d(f_m(s),f(s))$$

whenever $n,m \geq N$.

So far so good, now comes where my question lies. The proof then says:

Letting $m$ tend to $\infty$ we obtain $d(f_n(s),f(s)) < \epsilon$ for all $s \in S$. Consequently $\{f_n\}$ converges uniformly to $f$ on $S$.

-- What is the difference between $d(f_n(s),f(s))$ on the LHS and $d(f_m(s),f(s))$ on the RHS? i.e., if it's valid on the RHS, just let $n$ go to $\infty$ on the LHS and not need a proof.

-- Also where in the proof does the specificity "for each $s \in S$" which looks like pointwise convergence turn to uniform convergence?

Thanks

EDIT Maybe I should restate my question. The first term on the right in the long inequality $d(f_n(s),f_m(s))$ holds for all $s$ based on the definition of a Cauchy sequence of functions. So it comes down to the other two terms. In that there doesn't seem to be any inherent difference between them and the proof claims the one on the right becomes negligible for all $s$, then why could you not just start and end with that, using term $d(f_n(s),f(s))$ and not need the inequality at all.

And my second question is where exactly in the proof does it change from pointwise convergence? It was always the case for $d(f_n(s),f_m(s))$. But where does it change, e.g., for $d(f_m(s),f(s))$

-- What is the difference between $d(f_n(s),f(s))$ on the LHS and $d(f_m(s),f(s))$ on the RHS? i.e., if it's valid on the RHS, just let $n$ go to $\infty$ on the LHS and not need a proof

There is no difference between the two.

But if you let $n\to \infty$ in the RHS you get

$$0< \epsilon + d(f_m(s),f(s))$$

This is, of course, true but it does not yield any new information.

-- Also where in the proof does the specificity "for each $s \in S$" which looks like pointwise convergence turn to uniform convergence?

It becomes uniform convergence in the second part of the proof. To begin with you have a Cauchy sequence. The first part of the proof is to say what the limit is. For this you take the pointwise limit.

The second part is to show that this Cauchy sequence converges to that pointwise limit uniformly.

Edit

Yes, you are absolutely right. The proof could be shortened to this:

If for all $s \in S$ and $n,m \ge N$ we have $d(f_n(s), f_m(s)) \le \varepsilon$ then let $m \to \infty$ et voilà: $$d(f_n(s), f(s)) \le \varepsilon$$ as desired.

• Dear Rudy - Love the Ho Ho Ho
– user12802
Sep 12 '15 at 1:15
• @Andrew I tried to answer your edit but I don't understand the second question in the edit so I will answer the first for now. Sep 12 '15 at 1:43
• Actually you covered it in the edit as well. Thanks.
– user12802
Sep 12 '15 at 1:51

Maybe it would help if you choose $N_\epsilon$ so that $d(f_n(s), f_m(s)) < \epsilon/2$ for $n,m > N$. Then use the fact that $f_m \to f$ pointwise to choose $M_{\epsilon,s} > N$ so that $d(f_m(s), f(s)) <\epsilon/2$ for $m> M_{\epsilon, s}$. So $d(f_n(s), f(s) ) < \epsilon$ for $n > N_\epsilon$. This is uniform convergence since the final inequality depended on $\epsilon$ (not $s$).

The uniformity of the convergence is hidden in the definition of cauchy-sequence of the functions. We were able to choose $N_\epsilon$ such that $d(f_n(s), f_m(s))<\epsilon$ independent of/for all $s\in S$.