## Two-dimensional convergence

Here is something that I was certain would have an easy-to-prove positive answer; but it turns out that there is an exceedingly simple counterexample.

Let $\omega$ be the first infinite ordinal; its successor is $\omega + 1 = \omega \cup \{\omega\}$.We consider a map $M: (\omega+1) \times (\omega+1) \to \mathbb{R}$, or put differently, a $(\omega+1) \times (\omega+1)$-Matrix $M$ of real numbers.

We assume that $M$ has the following properties:
(1) For all $k\in \omega$ we have $\lim_{n\to\infty} M(k, n) = M(k,\omega)$, or more informally, we have “convergence to the right”.
(2) For all $k\in \omega$ we have $\lim_{n\to\infty} M(n, k) = M(\omega, k)$, or more informally, we have “convergence to the bottom”.
(3) $\lim_{n\to\infty} M(n,\omega) = M(\omega,\omega)$, or more informally, the right-hand entries of $M$ converge to the bottom right element $M(\omega,\omega)$.

Question: Does this imply that $\lim_{n\to\infty} M(\omega,n) = M(\omega,\omega)$, that is, do the bottom entries of $M$ also converge to  $M(\omega,\omega)$?

Answer: There is, surprisingly (to me, at least) an easy example showing that the answer is No. Let $M: (\omega+1)^2 \to \mathbb{R}$ be defined by $(\alpha,\beta)\mapsto 0$ if $\alpha \leq \beta$ and $(\alpha,\beta)\mapsto 1$ otherwise. It is easy to verify that (1), (2), (3) above are satisfied. Note that the right-hand entries $M(n, \omega)$ are all 0, and they trivially converge to $M(\omega,\omega) = 0$; but we have $M(\omega, n) = 1$ for all $n\in\omega$ therefore $\lim_{n\to\infty} M(\omega,n) = 1\neq 0 =M(\omega,\omega)$.

For the kind of two-dimensional convergence we are looking for we need some form of “simultaneous” convergence (conceptually related to uniform convergence), which I might address in a later post.