Let $F_{k,n} =\bigcap_{p,q \geq n}(x\in E,d(f_p(x),f_q(x))\leq \frac 1k ).$ let $E,F$ be a metric spaces such that $E$ is complete and $f_n$ a sequence of continuous maps of $E$ to $F$ we assume that $f_n$ converges. for all $k,n \in \mathbb N,k\geq 1,$ let $F_{k,n} =\bigcap_{p,q \geq n}(x\in E,d(f_p(x),f_q(x))\leq \frac 1k ).$
1)Verify that $\bigcup_{n\in \mathbb N}F_{k,n}=E$.
2)Conclude that $\Omega_k = \bigcup_{n\in \mathbb N}{F}_{k,n} ^o$ is open dense in $E$.
3) show that  for all $x \in \Omega_k$ has neighborhood $V$ such that $d(f(x),f(y))\leq \frac 3k$ for all $y\in V$.
My attempt:

*

*we have $F_{k,n}$ is closed. But how to use convergence to conclude that $\bigcup_{n\in \mathbb N}F_{k,n}=E$?.

*$\bigcup_{n\in \mathbb N}F_{k,n}=E$ and $E$ is complet space so by The baire category we have
$F_{k,n}^o\neq \emptyset$.

*$\Omega_k$ is open so there exists a neighberhood $V$ such that $V\subset \Omega_k$
 A: Let us show it step by step.
$\boldsymbol{1.}$ Let $x \in E$. As $f_n(x)$ converges in $F$ by assumption, $f_n(x)$ is a Cauchy sequence in $F$. Hence, for every $k$ in $\mathbb N_0$, we can find a $n$ sufficiently large so that $x \in F_{n, k}$, which implies that
$$E \subset \bigcup_{n} F_{n, k} \quad \Rightarrow \quad E  = \bigcup_n F_{n, k}~.$$
$\boldsymbol{2.}$ Clearly, $\Omega_k$ is open as an union of open sets. As you said, by Baire category theorem, there exists $n_0$ such that $F_{n_0, k}^o \neq \varnothing$. We denote $n_0 = \min\{n \in \mathbb N_0~|~F_{n, k}^o \neq \varnothing\}$. Note that $\bigcup_{n \ge n_0} F_{n, k} = E$ by the same argument as in $1.$ Moreover,
$$\text{cl} \left(\Omega_k\right) = \text{cl} \left(\bigcup_{n } F_{n,k}^o\right) \supseteq \bigcup_{n\ge n_0} F_{n,k} = E,$$
which proves that $\Omega_k$ is dense.
$\boldsymbol{3.}$ Exercise.
A: *

*has already been seen in the comments.


*If $E=\bigcup_{i\in\Bbb{N}} S_i$, where each $S_i$ is closed, we can write $S_i=\overline{S}_i=\mathop{\rm Fr}(S_i)\cup\mathring{S}_i$, where
$\mathop{\rm Fr}(S_i)=\big(\overline{S}_i\smallsetminus\mathring{S}_i\big)$ is the frontier of $S_i$.
Since frontiers are closed sets with empty interiors, thanks to Baire theorem the union $\bigcup_i\mathop{\rm Fr}(S_i)$ has also an empty interior.
But $E=\big(\bigcup_i\mathop{\rm Fr}(S_i)\big)\cup\big(\bigcup_i\mathring{S}_i\big)$; thus the open set $\bigcup_i\mathring{S}_i$ is dense.


*If $x\in\Omega_k$ then $x\in\mathring{F}_{k,n}$ for some $n\in\Bbb{N}$; let's put $U=\mathring{F}_{k,n}$.
We then have $d\big(f_p(x)\mathbin;f_q(x)\big)\leqslant\frac1k$ for all $p,q\geqslant n$.
Putting $p=n$ and letting $q\to\infty$ we obtain $d\big(f_n(x)\mathbin;f(x)\big)\leqslant\frac1k\cdot$
This inequality also holds for any $y\in U$, hence, by the triangle inequality:
\begin{eqnarray*}
\forall y\in U\quad d\big(f(x)\mathbin;f(y)\big)&\leqslant&d\big(f(x)\mathbin;f_n(x)\big)+d\big(f_n(x)\mathbin;f_n(y)\big)+d\big(f_n(y)\mathbin;f(y)\big)\\[2pt]
&\leqslant&d\big(f_n(x),f_n(y)\big)+\frac2k\cdot
\end{eqnarray*}
Now, since $f_n$ is continuous at $x$, there is some open set $V$ such that $x\in V\subset U$ and $d\big(f_n(x)\mathbin;f_n(y)\big)\leqslant\frac1k$ for all $y\in V$.
Conclusion: $d(f(x)\mathbin;f(y))\leqslant\frac1k+\frac2k\leqslant\frac3k$ for all $y\in V$.

