Does the completeness of a normed vector space only depend on its topology? 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 $(V, \Vert \cdot \Vert_2)$?
 A: 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.
A: 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.
A: 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.
A: 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.
