if two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?

Question

If two complete metric spaces coincide as sets, are their metrics (strongly) equivalent?

If $p$ and $q$ are two strongly equivalent metrics (in the sense that $ A p(x,y) \leq q(x,y) \leq B p(x,y)$ for some $A,B>0$) on some set $X$, then $(X,p)$ is complete if and only if $(X,q)$ is complete. My question is whether the converse is true as well, i.e. given two metrics $p$ and $q$ on $X$ such that $(X,p)$ and $(X,q)$ are complete, does it follow that $p$ and $q$ are strongly equivalent or at least equivalent (generating the same topology)?

If the answer is no, does the situation change in the context of Banach spaces, with metrics replaced by equivalent norms on some (of course infinite-dimensional) vector space?

In case the above converse is true, I guess one should argue by contradiction, assuming that the two metrics do not generate the same topology. Even in the case where the topology generated by one topology is strictly stronger than the other one, it is not clear to me how to find a sequence $(x_n)$ converging in the weaker metric (say $p$) but diverging in the stronger one.

Answer 1

That is not true. Let $f\colon\Bbb R\longrightarrow\Bbb R$ be any bijection. Let $p$ be the usual distance in $\Bbb R$ and define $q(x,y)=\bigl|f(x)-f(y)\bigr|$. Then both $(\Bbb R,p)$ and $(\Bbb R,q)$ are complete, but, in general, $p$ and $q$ are not equivalent.

Answer 2

Also fails for Banach spaces. Indeed, the non-isomorphic Banach spaces $l^1$ and $l^2$ are isomorphic as vector spaces; use that isomorphism to get two non-equivalent complete norms on the same vector space.

Remark
To get strong equivalence of two complete norms on one vector space, good tools are; closed graph theorem, open mapping theorem, Banach-Steinhaus theorem, and so on.

Answer 3

This is true in finite dimensional Banach spaces (since all norms on a given finite dimensional vector spaces are equivalent) but certainly not the case in the infinite dimensional case. In some sense, this is precisely the reason why topology plays such an important role in the general theory of Banach spaces.

Take any two significantly different Banach spaces, say $\ell_p$ and $\ell_q$ for $p\ne q$. The underlying sets are different, but they have the same cardinality. So, you can transport the structure from each Banach space to some fixed common set $X$. You now have two non-equivalent norms on $X$ inducing each a complete metric.