This is from Kurtz and Ethier's Markov Process.
For $n=1,2,\dots,$ let $L_n$, in addition to $L$, be a Banach space with norm also denoted by $\Vert \cdot \Vert$, and let $\pi_n:L\to L_n$ be a bounded linear operator. Assume that $\sup_n \Vert \pi_n\Vert <\infty$. If $A_n \subset L_n \times L_n$ is linear for each $n\ge 1$, the extended limit of the sequence $\{A_n\}$ is defined by $$\text{ex-}\lim_{n\to \infty}A_n=\{(f,g)\in L\times L: \text{there exists} \;(f_n,g_n)\in A_n \;\text{for each}\;n\ge 1 \;\text{such that}\;\Vert f_n - \pi_n f\Vert \to 0\;\text{and} \Vert g_n-\pi_n g\Vert \to 0\}.$$
Then ex-$\lim_{n\to \infty}A_n$ is closed in $L\times L$.
I have been unable to prove that this is closed. We need to show that given a sequence $(f_m,g_m)$ in the extended limit of the sequence that converges to some $(f,g)\in L\times L$, $(f,g)\in ex-\lim A_n$. But for each $f_m,g_m$ there is another associated sequence given by the definition of the set. How can I adjust these sequences to prove the desired result? I would greatly appreciate any help.