Are equivalence classes of norms that make the space complete unique? In my real analysis two class, we are talking about how $C([a,b],R)$ is the space of continuous functions from some real interval $[a,b]$ to the real numbers. We have talked about how a normed space can be considered complete if every Cauchy sequence of functions converges uniformly to a continuous function.
My question is:
If a normed space is complete wrt some norm. Can you also have a normed space using a norm not equivalent to the first for which the space will still be complete. When I say the word equivalent I am referring to two norms belonging to the same equivalence class. I am not saying that the norms are defined the same way.
Update: I am mainly interested in the space of continuous functions so any examples using those spaces would be ideal.
If anybody is still reading this look at the comment I left below and see if you can answer it. Thanks math experts!
 A: A similar question was asked a couple of months ago on Mathoverflow: can a field be complete with respect to two inequivalent absolute values?  Besides $\mathbf R$ and $\mathbf C$, there are many other fields complete with respect to a (nontrivial) absolute value: the $p$-adic numbers $\mathbf Q_p$, the "$p$-adic complex numbers" $\mathbf C_p$, the Laurent series fields $F((t))$ with the $t$-adic absolute value, and so on. While $\mathbf Q_p$ is locally compact, like $\mathbf R$ and $\mathbf C$, $F((t))$ is locally compact only when $F$ is a finite field and $\mathbf C_p$ is not locally compact.
Zorn's lemma (the axiom of choice) implies $\mathbf C$ and $\mathbf C_p$ are isomorphic as abstract fields, and using such a totally nonexplicit isomorphism to turn the absolute value on $\mathbf C_p$ into an absolute value on $\mathbf C$, we could say the complex numbers are complete with respect to an absolute value other than its usual one and with respect to which it is not locally compact. How strange! It is very nonconstructive. I don't think what you're seeking really occurs in any natural way.
In my answer here I show a field can't be locally compact for two inequvalent nontrivial absolute values.  On the same page, Arno Fehm points out an old theorem of F. K. Schmidt that shows if a field of characteristic $0$ is complete with respect to two inequivalent nonarchimedean absolute values then it must be algebraically closed.
So the fact that my example above uses $\mathbf C$ is not a surprise: Schmidt's theorem says any example in characteristic $0$ has to be algebraically closed.
A: No, they are not unique in general.
There is a nice answer by Daniel Fischer here. It is a bit more complicated than you would think by reading the comments above.
