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.

About dominiczypen

I'm interested in general topology, order theory, and graph theory. This link takes you to my preprints on arXiv.
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