I'm having trouble understanding what is meant with the following passage:
... functions s.t. $S(m,n)$ is defined whenever $m$ belongs to a directed set $D$, and $n$ belongs to a directed set $E_m$. We want to find a net $R$ with values in this domain s.t. $S\circ R$ converges to $\lim_m \lim_n S(m,n)$ whenever $S$ is a function to a topological space and this limit exists. It is interesting to note that the solution of this problem requires Moore-Smith convergence, for, considering double sequences, no sequence whose range is $\omega\times\omega$ can have this property.
(Quoted from "General Topology" by John K. Kelley, page 69)
Now what I don't quite understand is the remark about this not being possible when only considering 'normal' sequences. For example, if I set $S(n,m)=0\in\mathbb{R}$, $D=\mathbb{N}$, $E_m=\mathbb{N}$, then clearly $\lim_n\lim_m S(n,m)=0$. I can then just take $R:\mathbb{N}\to \mathbb{N}^2$ arbitrary and then always $\lim_n S\circ R=0$. So apparently I'm not really understanding what is meant here.
Now an other interpretation might be that using only sequences it is not possible to find a single sequence $R$ s.t. $S\circ R$ converges to the wanted limit, independent of the function $S$ (but of course all considered $S$ have the same domains). Now this I can imagine (I'm not sure if it's true), but for some reason it seems like a strange goal to want to find this single sequence $R$ for every possible $S$, so I'm not sure this is what is meant here.
By the way, for those curious: the wanted net $R$ is constructed by taking the product directed set $D'=D\times \prod E_n$ (with the standard induced order), and setting for $(m,f)\in D'$: $R(m,f)=(m,f(m))$.
What is meant is that there is no sequence $R:\omega\to\omega\times\omega$ such that for all functions $S$ with domain $\omega\times\omega$, $S\circ R$ converges to $\lim_m\lim_nS(m,n)$ whenever the codomain of $S$ is a topological space and the iterated limit exists. No matter what $R$ you try, there will be some space $X$ and some $S:\omega\times\omega\to X$ such that $\lim_m\lim_nS(m,n)$ converges to some $p\in X$, but either $S\circ R$ does not converge, or it converges to a point of $X\setminus\{p\}$.
In fact there are a single space $X$ and iterated sequence $S$ that ‘kill off’ every possible choice of $R$.
Let $X_0=\{\langle 2^{-m},2^{-n}\rangle:m,n\in\omega\}$, let $X_1=\{\langle 2^{-m},0\rangle:m\in\omega\}$, and let $p=\langle 0,0\rangle$. Let $Y=X_0\cup X_1$, and give $Y$ the topology that it inherits from the plane $\Bbb R^2$. For each function $\varphi:\omega\to\omega$ and $n\in\omega$ let
$$B(\varphi,n)=\{p\}\cup\{\langle 2^{-k},0\rangle:k\ge n\}\cup\{\langle 2^{-k},2^{-\ell}\rangle\in X_0:k\ge n\text{ and }\ell\ge\varphi(k)\}\;,$$
and take the family of all such sets $B(\varphi,n)$ as a local base at $p$. For future reference let $\varphi_0(n)=0$ for each $n\in\omega$. Let
$$S:\omega\times\omega\to X:\langle m,n\rangle\to\langle 2^{-m},2^{-n}\rangle\;;$$
clearly $\lim_nS(m,n)=\langle 2^{-m},0\rangle$ for each $n\in\omega$, and $\lim_m\langle 2^{-m},0\rangle=p$, so
$$\lim_m\lim_nS(m,n)=p\;.$$
Let $R:\omega\to\omega\times\omega$ be any sequence, and for each $n\in\omega$ let $A_n=\big\{k\in\omega:R(k)\in\{n\}\times\omega\big\}$. We distinguish two cases.
Suppose first that $A_n$ is infinite for some $n\in\omega$. Then $\langle (S\circ R)(k):k\in A_n\rangle$ is an infinite subsequence of $S\circ R$ lying entirely outside the nbhd $B(\varphi_0,n+1)$ of $p$, so $S\circ R$ cannot possibly converge to $p$.
Thus, $A_n$ is finite for each $n\in\omega$. Define $\varphi:\omega\to\omega$ by
$$\varphi(n)=\begin{cases} 1+\max A_n,&\text{if }A_n\ne\varnothing\\ 0,&\text{if }A_n=\varnothing\;; \end{cases}$$
then $B(\varphi,0)$ is a nbhd of $p$ disjoint from the range of $S\circ R$, so $S\circ R$ does not converge to $p$.
In other words, no matter what sequence $R:\omega\to\omega\times\omega$ we try, $S\circ R$ does not converge to $p$, even though $\lim_m\lim_nS(m,n)=p$.