Limit of limits converges

Question

Let $(a_k^\ell)_{k,\ell\in\mathbb N}$ be a double sequence that is Cauchy in the lower coordinate, i.e. $$\sup_{\ell\in\mathbb N}|a_m^\ell-a_n^\ell|<\varepsilon$$ for all $n,m\geq N(\varepsilon)$. I know that that $a_k:=\lim\limits_{\ell\rightarrow\infty} a_k^\ell$ exists for all $k$. Let's further assume that $a^\ell:=\lim\limits_{k\rightarrow\infty} a_k^\ell$ exists for all $\ell$ (i.e. we are in $c$-space).

I want to show that $(a^\ell)_{\ell\in\mathbb N}$ converges. My idea is to show it is Cauchy via an $\frac{\varepsilon}{3}$-argument as follows: $$|a^n-a^m|\leq|a^n-a_k^n|+|a_k^n-a_k^m|+|a_k^m-a^m|<\varepsilon$$

The problem is that the minimal $N(\varepsilon)$ is not well-defined in this situation, because the $k$ is dependent on $n$ and $m$, and $n$ and $m$ are dependent on $k$. Can I salvage this or do I have to prove that $(a_k)_{k\in\mathbb N}$ is convergent first?

Answer 1

Your approach can work. We just need to be careful about the order of applying our various limit definitions.

Fix $\varepsilon > 0$. Use your "uniformly Cauchy in the lower coordinate" assumption to choose $N = N(\frac \varepsilon 5)$ such that if $m,n \ge N$ then $|a_m^\ell - a_n^\ell| < \frac{\varepsilon}{5}$ for any $\ell$. Next, use existence of $\lim \limits _{\ell \to \infty} a_N^\ell$ to find $M \in \mathbb{N}$ such that if $k, \ell \ge M$ then $|a_N^k - a_N^\ell| < \frac \varepsilon 5$.

Finally, put the pieces together: if $k, \ell \ge M$ then we have

$$\begin{align} |a^k - a^\ell| &= |a^k - a_N^k + a_N^k - a_N^\ell + a_N^\ell - a^\ell| \\ &\le |a^k - a_N^k| + |a_N^k - a_N^\ell| + |a_N^\ell - a^\ell| \\ &\le \frac \varepsilon 5 + \frac \varepsilon 5 + \frac \varepsilon 5 \\ &< \varepsilon. \end{align}$$ The first and third terms are $\le \frac \varepsilon 5$ because of our choice of $N$, and the second term is $\le \frac \varepsilon 5$ by choice of $M$. We have thus shown the sequence $(a^\ell)$ is Cauchy, so it converges.

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To give a bit more intuition: this basically works out because the condition you called "Cauchy in the lower coordinate" is very strong! It basically says $a_n^\ell \to a^\ell$ as $n \to \infty$, uniformly in $\ell$. You can see this previous MSE question with more context and similar theorems. People there talk about a bunch of different conditions you could assume, but you can ctrl+F for uniform to find answers that use assumptions that are similar to yours.

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EDIT: More details on a step requested in comments

In a comment, OP asked for more detail on proving the following claim. Assume that for all $m,n \ge N$ we have $|a_m^k - a_n^k| < \varepsilon$. We claim that $|a_N^k - a^k| \le \varepsilon$. (Note to OP: In your comment you quote a version that uses $<$ instead of $\le$, but your version may not be true.)

To prove the claim, we fix some $\rho > 0$ find $R$ such that for $n \ge R$ we have $|a_n^k - a^k| < \rho$. Then let $T = \max\{N, R\}$. We have: $$\begin{align} |a_N^k - a^k| &= |a_N^k - a_T^k + a_T^k - a^k| \\ &\le |a_N^k - a_T^k| + |a_T^k - a^k| \\ &< \varepsilon + \rho. \end{align}$$ This proves $|a_N^k - a^k| < \varepsilon + \rho$ for any $\rho > 0$, which implies $|a_N^k - a^k| \le \varepsilon$ as desired.

If that last step still causes you trouble, you can prove it by contradiction. Assume $|a_N^k - a^k| > \varepsilon$. Then we could choose $\rho$ such that $0 < \rho < |a_N^k - a^k| - \varepsilon$, which would contradict the result established in the previous paragraph.