Completeness of Normed spaces. I want to prove the following proposition
If $(X,||\cdot||)$ and $(X,||\cdot||')$ are homeomorphic, then $(X,||\cdot||)$ is complete if and only if $(X,||\cdot||')$ is complete.
So, I only know the definition of homeomorphic, and can't figure out how only with that I can prove that proposition. Can someone help me to prove this please?
Thanks a lot in advance :) 
 A: As a sketch$\ldots$
Well if you've a homeomorphism $f$, then if $\{x_n\}_{n \in \Bbb N}$ is cauchy wrt $\|\cdot \|$, and converges to $x \in X$, then $\|x_n-x\|<\delta$ for $n$ sufficiently large, and $\|x_n-x_m\|<\delta$, for $n,m$ sufficiently large. Since $f$ is continuous we can make $\|f(x_n)-f(x_m)\|'<\epsilon$ for $\|x_n-x_m\|<\delta$, so $\{f(x_n)\}_{n \in \Bbb N}$ is cauchy. Now since $f$ is continuous at $x$ we also know that $\|f(x_n)-f(x)\|'<\epsilon$ for $\|x_n-x_m\|<\delta$. So $f(x_n)$ will converge to $f(x)$. Then to prove the other way, you do the same thing, except with $f^{-1}$, which is continuous since $f$ is a homeomorphism.
A: Take a Cauchy sequence in $(X,||\cdot||)$ and show that is also a Cauchy sequence in $(X,||\cdot||')$.
This proves the assertion since convergence is a topological property which is preserved by homeomorphisms.
A: Since $(X, ||\cdot || ) $ and $ (X, ||\cdot ||' )$ are homeomorphic the topologies generated by these norms are identical. Hence the identity map $\mbox{Id} :(X, ||\cdot || ) \to (X, ||\cdot ||' )$ is continiuous so the norms are equivalent. And from the equivalence of the norms it easily follows your assertion.
