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Let $O(n)$ to be the orthogonal group, and define $m(A)$ for every $A \in O(n)$ as follows: $$m(A) = \max \left\{ \frac{|Ax-x|}{|x|} : x \in \mathbb{R} ^ n -\{0\} \right\}\;.$$ Given a sequence $ \{B_i \}$ , we can assume by compactness that it converges.

Why does this imply that for every $\epsilon>0$ there exist $i<j$ such that $m(B_j B_i^{-1} )\leq \epsilon $ ?

Thanks in advance

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up vote 2 down vote accepted

Suppose $B_i \to L$ as $i \to \infty$. Since both multiplication and inverse are continuous, as $i \to \infty$ and $j \to \infty$ we have $B_j B_i^{-1} \to L L^{-1} = I$. Now $m$ is continuous (note that by homogeneity we may take $x$ to be in the unit sphere which is compact), and $m(I) = 0$.

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