Metric and Topological structures induced by a norm
Solution 1:
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No. It is straightforward to show that equivalent norms yield both the same convergent sequences and the same Cauchy sequences. (Written before Rasmus's answer was posted, but posted afterward.)
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Yes. One way to see this is to note that isomorphism classes of vector spaces depend only on linear dimension, so the question amounts to finding 2 nonisomorphic Banach spaces of the same linear dimension. There are lots of examples of these. Every infinite dimensional separable Banach space has linear dimension $2^{\aleph_0}$. However, for example, $\ell^1$ and $c_0$ are separable Banach spaces that are not isomorphic (as Banach spaces).
Actually, "amounts to" wasn't quite accurate. It is certainly sufficient that the 2 Banach spaces are not isomorphic, but it is not necessary because you are only asking that one particular map (the identity in the original formulation) is not an isomorphism. So what I gave above is actually stronger. To just answer 2), you could just take any infinite dimensional Banach space and induce a new norm via an unbounded linear isomorphism with itself.
The answer above was assuming that equivalent norms are defined as in this PlanetMath article. If instead you meant only that the spaces are homeomorphic in the norm topologies, as Jyotirmoy Bhattacharya suspected, then the examples alluded to above won't work. However, there are also examples of pairs of Banach spaces that have the same linear dimension but are not homeomorphic, and this will work in either case. For example, $\ell^\infty$ and $c_0$ are not homeomorphic because $\ell^\infty$ is nonseparable. Both spaces have linear dimension $2^{\aleph_0}$. This was already mentioned for $c_0$, and for $\ell^\infty$ it follows because $c_0$ embeds in $l^\infty$ (which gives the lower bound on dimension) and because the cardinality of $\ell^\infty$ is $2^{\aleph_0}$ (which gives the upper bound).
(I'm now pretty sure this isn't what you want, based on your edit, but this still gives another example for the actual question as well as an answer to Jyotirmoy's comment.)
Incidentally, another way to see that $2^{\aleph_0}$ is a lower bound for the linear dimensions of $\ell^1$ and friends is to consider the linearly independent set $\\{(1,t,t^2,t^3,\ldots):0\lt t\lt 1\\}$. Cardinality of the spaces gives an upper bound.