Let $M$ be the space of $n\times n$ real matrices. It becomes a metric space when endowed with the inner product $g(A,B)=\text{tr}(A^T B)$. Let $d$ be the induced metric.
Let $F: M \times M \rightarrow M$, $F(A,B)=AB$.
Is the matrix product $F$ continuous, with respect to the product topology and the topology induced by $d$?
My attempt: let $\mathbf{X},\mathbf{Y} \in M^2$.
I want to show that $d_p(\mathbf{X},\mathbf{Y})=\sqrt{d(X_1,Y_1)^2+d(X_2,Y_2)^2}$ metrizes the product topology. If so the continuity proof becomes a standard $\epsilon - \delta$ argument.
From the fact that $d$ is a metric $d_p$ is also a metric.
To see it induces the product topology, let $d_{\infty}(\mathbf{X},\mathbf{Y})=\max \{d(X_1,Y_1),d(X_2,Y_2)\}$. It is a metric as well and equivalent to $d_p$.
Now, let $B_{d_{\infty}}(\mathbf{X},r)$ be the open ball of radius $r$ about $\mathbf{X}$ (w.r.t. $d_{\infty}$). $\mathbf{Y} \in B_{d_{\infty}}(\mathbf{X},r) \iff \max \{d(X_1,Y_1),d(X_2,Y_2)\} <r \iff d(X_i,Y_i)<r, i=1,2 \iff \mathbf{Y}\in B_d(X_1,r) \times B_d(X_2,r)$.
It follows that $B_{d_{\infty}}(\mathbf{X},r)=B_d(X_1,r) \times B_d(X_2,r)$, so my claim follows.
Have I committed any mistake? Is there any less painful way of proving continuity of the matrix product? Thanks!
There is a simple way of seeing that that function is continuous. Since, in a finite dimensional real vector space, all norms are equivalent, you can assume, without loss of generality, that you are dealing with the Euclidean distance:$$d\bigl((a_{ij})_{1\leqslant i,j\leqslant n},(b_{ij})_{1\leqslant i,j\leqslant n}\bigr)=\sqrt{\sum_{i,j=1}^n{a_{ij}}^2}.$$Besides, a function $f$ from a topological space into $\mathbb R^N$ is continuous if and only if each of its $N$ components is continuous. But each component of the product of two sequare matrices is a polynomial in the entries of the matrix. Therefore, it is continuous.