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Let $V \space$ be a vector space over $\mathbb{R}$, and $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ norms over $V$, which generate the same topology. Is it always true that if $v_n$ is a Cauchy sequence with respect to both norms, and $v_n$ converges in $(V, \Vert \cdot \Vert_1)$ then it converges in $(V, \Vert \cdot \Vert_2)$? If $dim(V)<+\infty$ the assertion is true, because there exist constants $c,C\in\mathbb{R}$ such that $\forall v\in V \space$ $c\Vert v \Vert_1 \leq \Vert v \Vert_2 \leq C\Vert v \Vert_1$, but I have a feeling it isn't in general (this would be strange, since completeness isn't a topological property; however maybe the additional structure of vector space might be used in some way).

EDIT
Thanks to everyone's answers I have realized that the above question was badly stated to begin with.
What I meant to ask was whether the property of a normed vector space of being complete only depends on the topology generated by the norm, and not by the norm itself. As stated indirectly by many users, convergence, unlike the property of being a Cauchy sequence, is a topological property. Therefore the answer to the question I actually asked is always affermative.
What I should have asked was: given two topologically equivalent norms $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ on a vector space V, is it true that a sequence $v_n$ is a Cauchy sequence in $(V, \Vert \cdot \Vert_1)$ if and only if it is in $(V, \Vert \cdot \Vert_2)$?

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    Excuse me, but when $||\cdot||_1$ and $||\cdot||_2$ generate the same topology on $V$, being the convergence a topological property, for any sequence $v_n$, we have that $v_n$ is convergent w.r.t. the first norm iff $v_n$ converges w.r.t. to the second one, and in the affirmative case the limits are the same.2011-11-28
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    Yes, that is exactly what I meant. I was looking for a counterexample in the infinite dimensional case.2011-11-28
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    I would remark that the question in your second paragraph has an obvious positive answer. You should look more closely to the question if there exist two norms on a vector space(necessarily infinite dimensional) which generate the same topology but are not equivalent.2011-11-28
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    Ok sorry...I guess I mixed things up a bit. Here is what I meant. If $dim(V)<+\infty$ then I know that one normed space is complete if and only if the other one is. In the infinite dimensional case, however, this isn't so obvious (at least not to me) since I can't rely on the equivalence of the two norms to prove convergence. As you said, finding two non-equivalent norms which generate the same topology could be a good starting point, but still wouldn't answer the question. If I prove that all norms which generate the same topology are equivalent, then the question is aswered affermatively.2011-11-28
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    The two norms $||\cdot||_1$ and $||\cdot||_2$ on the vector space $V$ generate the same topology iff and only there exist real positive numbers $r$ and $R$ s.t. $||x||_1, and $||x||_2, but this can easily rewritten as $r||x||_2\le ||x||_1$ and $R||x||_1\le ||x||_2$ (that is the equivalence condition for oour norms).2011-11-28

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