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let $A,B$ be $m \times n$ matrices . $\|A\|$ := the square root of sum of (individual entry square) (hope it's clear :P) $d(A, B) = \|A − B\|$, already proved that $d$ is a metric.

(1)now proved that matrix addition is continuous under this metric topology $+ : M(m, n) \times M(m, n) \to M(m, n)$. You may use the fact that addition on $\mathbb{R}$ is continuous (this can be proven, e.g., using the epsilon− δ definition). (2) Prove that matrix multiplication $× : M(m, n) \times M(n, r) \to M(m, r)$ is continuous under this metric topology. You may use the fact that addition and multiplication on $\mathbb{R}$ are continuous (again, these can be proven, e.g., using the $\epsilon$-$\delta$ definition).

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$\|(A+B)-(C+D)\| = \|(A-C)+(B-D)\| \le \|A-C\| + \|B-D\|$. Hence if $\|A-C\|, \|B-D\| < {1 \over 2} \epsilon$, then $\|(A+B)-(C+D)\| < \epsilon$.

$\|AB-CD\| = \|AB-CB+CB-CD\| \le \|AB-CB\|+\|CB-CD\|$. Now suppose $\|A\|,\|B\| < M$ and $\|A-C\|, \|B-D\| < \min(1,{1 \over 2(M+1)} \epsilon)$, then $\|AB-CD\| \le \|B\| \|A-C\| + \|C\| \|B-D\| < \epsilon $. (This uses the fact that the Frobenius norm is submultiplicative.)

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