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I have a previous similar question. I'm working out that one with the answerer, but I'm trying to gain insight from a different angle, especially in approaching these problems.

I must establish that two vectors $A$ and $B$ are equal asymptotically (length of vectors, $n\to\infty$). I consider the error vector, $e=A-B$ and try to show that $e\to 0$ as $n\to\infty$.

I then show that for each element $e_i$ of the error vector, $\Vert e_i\Vert_2\to 0 $ as $n\to\infty$. How can I proceed from here? Is this sufficient to say that the two vectors are equal in some sense?

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Not sure I quite follow, but suppose $A$ is all zeros except for a one in its last component, and $B$ is all zeros except for a one in its next-to-last component. Then each component of the difference goes to zero but the difference doesn't go to zero as the number of components goes to infinity.

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    _"Then each component of the difference goes to zero"_... I don't follow. In this case, won't $\Vert e_{\text{last}-1}\Vert_2=\Vert e_{\text{last}}\Vert_2=1$? I can definitely show that $\Vert e_i\Vert_2\to 0$ for all $i$ in my case.2011-12-27
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    @steve: Let $e_i(n)$ be the $i$th component of the difference at the $n$th step. In Gerry Myerson's example, $\lVert e_i(n)\rVert\to 0$ as $n\to\infty$ for each *fixed* $i$. While it is true that $\lVert e_{n-1}(n)\rVert = \lVert e_n(n)\rVert=1$, you are changing the component as you change $n$ when you do that evaluation ("last" and "last-1" are not constant).2011-12-27
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    @ArturoMagidin I see. Thanks for clarifying. I guess I will need to show that $\Vert e \Vert\to 0$ and not component wise.2011-12-27