A Diagonalization Argument Involving Double Limits

For simplicity, write $(a_n)_{n\in\mathbb{N}} \subseteq (b_n)_{n\in\mathbb{N}}$ if $(a_n)_{n\in\mathbb{N}}$ is a subsequence of $(b_n)_{n\in\mathbb{N}}$. Then we have:

Lemma. For each $a \in \mathbb{N}$ and $(n_k)_{k\in\mathbb{N}} \subseteq (n)_{n\in\mathbb{N}}$, there exists $(\tilde{n}_k)_{k\in\mathbb{N}} \subseteq (n_k)_{k\in\mathbb{N}}$ such that

$$\lim_{k\to\infty} f_{k,\tilde{n}_{k}}(a) = f(a) $$

Proof. We recursively define $(\tilde{n}_k)_{k=1}^{\infty}$ as follows:

• Write $\tilde{n}_0 = 0$ for brevity, although this will not be included in the sequence $(\tilde{n}_k)_{k=1}^{\infty}$.

• If $\tilde{n}_{k-1}$ is defined, then choose $\tilde{n}_k \in (n_j)_{j\in\mathbb{N}}$ so that $\tilde{n}_k > \tilde{n}_{k-1}$ and $|f_{k,\tilde{n}_k}(a) - A_k(a)| < 2^{-k}$.

Here, $A_k(a) = \lim_{n\to\infty} f_{k,n}(a)$ is as in OP, and this explains why we can choose such $\tilde{n}_k$'s. By construction, it is clear that $f_{k,\tilde{n}_k}(a) \to f(a)$ as $k\to\infty$, proving the desired claim. $\square$

Now we may apply lemma for each $a = 1, 2, 3, \cdots$ to obtain $(n^{1}_k)_{k\in\mathbb{N}} \supseteq (n^{2}_k)_{k\in\mathbb{N}} \supseteq (n^{3}_k)_{k\in\mathbb{N}} \supseteq \cdots $ such that $f_{k,n^a_k}(a) \to f(a)$ as $k\to\infty$ for each $a$. Then we can apply diagonalization argument to construct $(n'_k)_{k\in\mathbb{N}}$, given by $n'_k = n_k^k$, so that

$$ \forall a \in \mathbb{N}, \qquad \lim_{k\to\infty} f_{k,n'_k}(a) = f(a). $$