A Diagonalization Argument Involving Double Limits The following problem is confusing me a lot:
Suppose that $\mathbb{N}$ and $\mathbb{R}$ denote the sets of positive integers and real numbers, respectively, and for each $(m,n) \in \mathbb{N}\times \mathbb{N}$, let $f_{m,n} : \mathbb{N} \mapsto \mathbb{R}$ be a function. Suppose that for every $a\in \mathbb{N}$, $$\lim_{m\rightarrow \infty} \lim_{n \rightarrow \infty} f_{m,n}(a) = f(a)$$ for some function $f:\mathbb{N}\mapsto \mathbb{R}$. Note that, the above equation means that for every $a \in \mathbb{N}$ and every $m \in \mathbb{N}$, $A_m(a) := \lim_{n \rightarrow \infty} f_{m,n}(a)$ exists, and for every $a \in \mathbb{N}$, $A_m(a) \rightarrow f(a)$ as $m \rightarrow \infty$.
My question is, does there exist two sequences of positive integers $\{m_j\}_{j=1}^\infty$ and $\{n_j\}_{j=1}^\infty$, such that:
$$\lim_{j\rightarrow \infty} f_{m_j,n_j}(a) = f(a)$$ for every $a \in \mathbb{N}$? I was trying to use a diagonalization argument, but I am getting more and more confused! In case my claim is not true, a counterexample would be nice.
Any help will be greatly appreciated.
 A: 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
$$ \forall a \in \mathbb{N}, \qquad \lim_{k\to\infty} f_{k,n'_k}(a) = f(a). $$
A: Nice answer Sangchul Lee!
By the way, I worked out the following solution to my own question, which uses a kind of reverse diagonalization argument. Can anyone comment on this solution (whether I am correct or not)?
The double limit condition clearly guarantees the existence of $m_1, n_1 \in \mathbb{N}$, such that $|f_{m_1,n_1}(1) - f(1)| < 1$. Assume inductively, that $m_1<m_2<\cdots<m_{k-1}$ and $n_1<n_2<\cdots<n_{k-1}$ have been defined. Clearly, there exists $m_k > m_{k-1}$ such that $$\Big|\lim_{n\rightarrow\infty} f_{m_k,n}(i) - f(i)\Big| < \frac{1}{2k}$$ for all $i \in \{1,2,\cdots,k\}$. Now, there exists $n_k > n_{k-1}$, such that $$\Big|f_{m_k,n_k}(i) - \lim_{n\rightarrow\infty} f_{m_k,n}(i)\Big| < \frac{1}{2k}$$ for all $i \in \{1,2,\cdots,k\}$. Hence, by triangle inequality,
$$\Big|f_{m_k,n_k}(i) - f(i)\Big| < \frac{1}{k}$$ for all $i \in \{1,2,\cdots,k\}$. The sequences $\{m_k\}$ and $\{n_k\}$ are constructed inductively in this way, and satisfy my claim.
Can you please confirm, whether I am correct or not? Thanks in advance!
