Let $\phi : E \to E'$ be a bi-Lipschitz map between metric spaces. Show that $E$ is complete if and only if $E'$ is complete.
I'm unsure of how to construct this proof. I know from a previous exercise that if $(x_n)$ is a Cauchy sequence in $E$, then $(\phi(x_n))$ is a Cauchy sequence in $E'$, if required.
Many thanks!
Since $\phi$ is bi-Lipschitz, then you know that its inverse $\phi^{-1}\colon E' \to E$ is Lipschitz continuous. In particular you have that $$ (1) \qquad (x_n)\subset E\ \text{Cauchy} \quad \Longrightarrow \quad (\phi(x_n)) \subset E' \ \text{Cauchy} $$ and $$ (2) \qquad (x'_n)\subset E'\ \text{Cauchy} \quad \Longrightarrow \quad (\phi^{-1}(x'_n)) \subset E \ \text{Cauchy} $$ Let us assume that $E$ is complete. Let $(x'_n)\subset E'$ be a Cauchy sequence. By (2) we have that $(\phi^{-1}(x'_n))$ is a Cauchy sequence in $E$. Since $E$ is complete, there exists $x\in E$ such that $\phi^{-1}(x'_n) \to x$, hence $\phi(\phi^{-1}(x'_n)) \to x':= \phi(x)$. Hence any Cauchy sequence in $E'$ is convergent.
In a similar way, using (1), you can prove the other implication.