Is this operation on infinite matrices associative?

Question

This is an additional question to that question, inspired by Leonid Petrov.

Let $\text{Mat}(\mathbb{N},\{0,1\})$ be the set of all maps $A:\mathbb{N}\times\mathbb{N}\to \{0,1\}$. We define a matrix multiplication for $A, B\in \text{Mat}(\mathbb{N},\{0,1\}$) and $m,n\in\mathbb{N}$ by $$(A B)(m, n) = 0 \text{ if } \sum_{i=1}^\infty A(m, i)B(i, n) < \infty \text{ and } (AB)(m, n) = 1 \text{ otherwise.}$$

Is this kind of matrix multiplication associative?

Answer 1

No, even $(A^2)A\ne A(A^2)$ in general. We consider $A$ as an incidence matrix of an infinite graph (with possible loops, and directed, but in our example it is loop-less and undirected). Consider vertices 1, 2 and join them but infinitely many paths $1x_{ij}y_j2$, where $i,j=1,2,\dots$, and all vertices $y_j,x_{ij}$ are distinct. Then $(A^2)_{1y_j}=1$ for all $j$, hence $((A^2)A)_{12}=1$, but $(A^2)_{k2}=1$ only for $k=2$, thus $(A(A^2))_{12}=0$.