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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. –  Giuseppe Tortorella Nov 28 '11 at 15:27
    
Yes, that is exactly what I meant. I was looking for a counterexample in the infinite dimensional case. –  Emilio Ferrucci Nov 28 '11 at 15:33
    
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. –  Giuseppe Tortorella Nov 28 '11 at 15:40
    
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. –  Emilio Ferrucci Nov 28 '11 at 16:20
    
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. $||x||_1<r\Rightarrow ||x||_2<1$, and $||x||_2<R\Rightarrow ||x||_1<1$, 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). –  Giuseppe Tortorella Nov 28 '11 at 16:27

4 Answers 4

I think you are mixing things up. The definition of equivalence of norms is what you said: there exists $c,C\in\mathbb{R}$ such that $c\|v\|_1≤\|v\|_2≤C\|v\|_1$, and so $V$ will be complete with respect to one norm if and only if it is complete with respect to the other. Therefore, completeness is a topological property (see Julian comment).

However, we have to assume $V$ is finite-dimensional in order to conclude that all norms are equivalent.

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Completeness is a metric property. For instance, $(-1,1)$ and $\mathbb{R}$ are homeomorphic with the topology induced by the usual metric (an homeomorphism is $\phi(x)=x/(1-|x|)$,) but $(-1,1)$ is not complete, while $\mathbb{R}$ is. To understand what is happening consider the sequence $\{1-1/n\}\subset(-1,1)$. It is a Cauchy sequence, but does not converge in $(-1,1)$. The sequence $\{\phi(1-1/n)\}=\{n-1\}$ is unbounded, and in particular is not a Cauchy sequence. –  Julián Aguirre Nov 28 '11 at 15:44
    
Thank you for the answer and the comment. @M Tureon I understand that in the finite dimensional case completeness (or rather, since we are dealing with vector spaces, "banachness") is a topological property. I was wondering what happens in the infinite dimensional case. @Julian What you said is the reason I asked the question in the first place. I was looking for a counterexample to show that completeness depends on the metric (in this case, the norm) and not only the topology, in the particular context of the question. –  Emilio Ferrucci Nov 28 '11 at 15:53

If the two norms give rise to the same topology, then the identity map is an homeomorphism, ans since it's a linear map, those two norms are equivalent. And so they are simultaneously complete.

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How do you conclude that the two norms are equivalent, using the fact that the identity is a linear map? –  Emilio Ferrucci Nov 28 '11 at 15:59

Let V be a normable space (which fixes a topology $T$ on V), and $\| \cdot \|_1$ and $\| \cdot \|_2$ be two norms that induce the topology on $V$, then the identity map $$ id: (V, \| \cdot \|_1) \to (V, \| \cdot \|_2) $$ is a homeomorphism (as is its inverse). This implies that the unit ball with respect to $ \| \cdot \|_2 $ contains a multiple of the unit ball with respect to $\| \cdot \|_1$ (and vice versa), ergo the two norms are equivalent. There is no need for any algebraic structure on V.

The completeness of a topological vector space is a concept that does depend on the topology only: A topological vector space V is complete if every Cauchy filter converges to a point in V. This is independent of the question if V is normable and - if it is - which norms induce the topology on V.

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This is essential property of normed spaces. Completeness is an invariant of isomorphism in the category of normed space, not metric! For example metric spaces $(\mathbb{R},d_1)$, $(\mathbb{R},d_2)$, where $d_1(x,y)=|x-y|$, $d_2(x,y)=|e^x-e^y|$ are homeomrphic but the first is complete and the second does not complete.

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