"composition" of "pointwise convergent sequences of functions" Intuitively, if $f_n\to f$ as $n\to\infty$ and $g^{(n)}_i\to f_n$ as $i\to\infty$, can we get $g_j\to f$ as $j\to\infty$?
Formally,

Let $\{f_n\}_n$ be a sequence of functions from $\mathbb{R}^d$ to
  $\overline{\mathbb{R}}$, the extended real line. Let $f$ be its
  pointwise limit, i.e. for each $x\in\mathbb{R}^d$ and each
  $\varepsilon>0$, there exists $N\in\mathbb{Z}^+$ such that
  $|f_n(x)-f(x)|<\varepsilon$ for all $n\geq N$. Each $f_n$ is the
  pointwise limit of sequence $\{g^{(n)}_i\}_i$. Can we construct a
  sequence of $g$'s converging pointwise to $f$? If not, what additional
  conditions do we need?

As usual, any help is appreciated:)
 A: An old question, but maybe someone likes to read about an answer somewhere in the future.
This is a basic fact about topology, the property you are asking for is an axiom of a topology defined through nets (generalized sequences).
So in a topological space $(X,\tau)$, we assume that around $x\in X$ we a have a countable basis of neighbourhoods and we take a converging sequence $x_n\to x$ and a sequence of converging sequences $y_m^n\to x_n$ for all $n\in\mathbb{N}$.
Now for any open neighbourhood $U$ of $x$ we have that there exists $N_U\in\mathbb{N}$ such that
$$
x_n\in U\quad\forall n>N_U.
$$
Since $U$ is open we get a natural number $M^n_U$ for all $n>N_U$ such that
$$
y^n_m\in U\quad\forall m>M^n_U.
$$
So now by taking $U_k$ a basis for the neighbourhoods of $x$, for any $k\in\mathbb{N}$ we can just take any
$$
y^n_m\in U_k\quad  m>M^n_{U_k},\, n>N_{U_k},
$$
and get the desired sequence convergin to $x$.
Given the set of functions $\mathrm{Set}(\mathbb{R}^d,\overline{\mathbb{R}})$, the pointwise convergence is the finest topology that makes all the evaluations continuous, and thus you can apply the above to the topological space $\mathrm{Set}(\mathbb{R}^d,\overline{\mathbb{R}})$.
