# Equicontinuous sequence of functions that converges pointwise in a dense subset of its domain, converges in the whole domain.

The original exercise had the aditional hypothesis that the codomain was complete. I think it's not necessary:

Let $f_n:E\rightarrow F$ be the sequence in question and let $D\subseteq E$ be the dense subset. Take $a$ in E. Let $\epsilon >0$ be given. Because of equicontinuity, there exists $\delta > 0$ such that $d(x,a) < \delta$ implies $d(f_n(x),f_n(a)) < \dfrac{\epsilon}{2}$, for every $n$ in $\mathbb{N}$. Let $y$ be a point in $B(a,\delta) \cap D$. By hypothesis, $f_n(y)$ converges to some value, call it $u$. Then:

$d(f_n(a),u)\le d(f_n(a), f_n(y))+d(f_n(y),u)$. For $n\ge N$, we have $d(f_n(y),u)<\dfrac{\epsilon}{2}$. Then $f_n(a)$ converges to $u$. My problem is the following: This proof implies that for every $x$ in $B(a,\delta) \cap D$, $f_n(x)$ converges to the same value $u$. Also, If I take some $b$ in $B(a,\delta)$, it can also be seen that there is some ball with center in $b$ in which at the same time equicontinuity applies and is contained in $B(a, \delta)$, which implies that $f_n(b)$ converges to $u$ too. So, this would imply that given $a$ in E, $f=\lim f_n$ is constant in some ball with center $a$. If we think now in $E=F=\mathbb{R}$, $f$ must be constant, because the pointwise limit of equicontinuous functions has to be continuous. Where is my flaw?

You need to use completeness to show that any sequence $(f_n(a))$ converges for $a \in E$.

Take any $a \in E$.

Since $D$ is dense in $E$, there exists $b \in D$ such that $d(b,a) < \delta$ and $d(f_n(b), f_n(a)) < \epsilon/3$ for all $n$.

We have for $m > n$

$$\tag{*}d(f_m(a),f_n(a)) \leqslant d(f_m(a),f_m(b)) + d(f_m(b),f_n(b)) + d(f_n(b),f_n(a)) \\< \frac{2\epsilon}{3} + d(f_m(b),f_n(b))$$

Since $(f_n(b))$ is convergent we can show using (*) that $(f_n(a))$ is a Cauchy sequence.

I try an explanation

In fact, you have two $\varepsilon$. The first, say $\varepsilon_1$, is used when you say that we have $d(f_n(x),f_n(a))<\varepsilon_1/2$ for $d(x,a)<\delta(\varepsilon_1)$. Then you have obtained that for a fixed $y$ in the dense subset, with $f_n(y)\to u$, such that $d(y,a)<\delta(\varepsilon_1)$, you have $d(f_n(a),u) \leq d(f_n(y),f_n(a))+d(f_n(y),u)\leq \varepsilon_1/2 +d(f_n(y),u)$; you cannot now change $\varepsilon_1$, as $y$ is fixed. Now, for every $\varepsilon_2>0$, you can find an $N$ such that for $n\geq N$, you have $d(f_n(a),u)\leq \varepsilon_1/2+\varepsilon_2/2 =\varepsilon$. But $\varepsilon\geq \varepsilon_1/2$, so it is not abritrary. You have not obtained the convergence of $f_n(a)$. You have only obtained that if $f_n(a)\to v$, then $v$ is close to $u$.