Let $\omega$ be the first infinite ordinal, and let $A$ be a real $(\omega+1)\times(\omega+1)$-matrix, that is $A$ is a map $A:(\omega+1)\times(\omega+1) \to \mathbb{R}$.


Suppose that $A$ has the following properties:


1. for all $k\in\omega$ we have $\lim\_{n\to\infty}A(k, n) = A(k,\omega)$, or, in other words, the entries of every horizontal line converge to the rightmost element of that line;
2. for all $k\in\omega$ we have $\lim\_{n\to\infty}A(n,k) = A(\omega,k)$, or, in other words, the entries of every vertical line converge to the bottom element of that line;
3. $\lim\_{n\to\infty} A(n,\omega) = A(\omega,\omega)$, or, in other words, the right-hand elements of the matrix converge to the right bottom element.


Does this imply that $\lim\_{n\to\infty}A(\omega, n) = A(\omega,\omega)$, that is the bottom line also converges to $A(\omega,\omega)$, the rightmost bottom element?



Convergence in an infinite matrix
No - there is even a counterexample with just entries in $\{0,1\}$.


Define $A:(\omega+1)^2\to \mathbb{R}$ by $(\alpha,\beta)\mapsto 0$ if $\alpha\leq \beta$, and $(\alpha,\beta)\mapsto 1$ otherwise. That is $A$ is the $(\omega+1)\times(\omega+1)$-matrix such that the elements on the main diagonal (and above) are $0$, and the other ones are $1$.


It's not hard to verify that the three conditions of the questions are satisfied. In particular note that the "right-hand entries" $A(n,\omega)$ are $0$ and they trivially converge to $A(\omega,\omega)=0$.


However, in the bottom line we have $A(\omega,n) = 1$ for all $n < \omega$, but $A(\omega,\omega) = 0$, therefore $\lim\_{n\in\omega}A(\omega,n) = 1\neq A(\omega,\omega)$.


