Prove a theorem involving normed spaces and completeness. Theorem:
If $(X,||\cdot||)$ and $(X,||\cdot||')$ are homeomorphic, then $(X,||\cdot||)$ is complete if and only if $(X,||\cdot||')$ is complete.
So my goal is to prove that since they are homeomorphic then they have the same open sets, so I want to apply the following result 
Let $||\cdot||_1$ and $||\cdot||_2$ two norms on $X$, and suppose there exist $\delta$, $\epsilon>0$ such that $B^1_\delta(0) \subset B_1^2(0)$ and $B_\epsilon^2(0)\subset B_1^1(0)$. Then $||\cdot||_1$ and $||\cdot||_2$ are equivalent.
So I can have equivalence of norms:
Two norms $||\cdot||_1$ and $||\cdot||_2$ on a vector space are equivalent if there exist constants $c_1$, $c_2$ such that
$$c_1||x||_1 \le ||x||_2 \le c_2||x||_1$$
for all $x\in X$.
And therefore I can conclude the result, but How can I prove with the above result that I get equivalent norms in this theorem?.
Thanks a lot in advance :).
 A: Notice that for $x \neq 0$,$$\frac{\delta}{2}\frac{x}{\|x\|_1} \in B^1_{\delta}(0) \subset B^2_1(0) \quad \Rightarrow \quad \Big\|\frac{\delta}{2}\frac{x}{\|x\|_1}\Big\|_2 \le 1 \quad \Rightarrow \quad\|x\|_2 \le \frac{2}{\delta}\|x\|_1.$$
The other inequality is obtained similarly, using the other known inclusion. 
Clearly if $x = 0$ the inequality is trivially satisfied.


Addendum#1: $X_i$ refers to $(X,\|\cdot\|_i)$. Assume that $X_1$ is complete and consider a Cauchy sequence, $\{x_n\}$, in $X_2$.
By the equivalence of the norms it follows that $\{x_n\}$ is a Cauchy sequence in $X_1$ as well (which is complete by assumption) and hence it is convergent to a certain $x \in X$. Indeed, $$\|x_n - x_m\|_1 \le \frac{1}{c_1}\|x_n - x_m\|_2 \to 0.$$But now, using the other branch of the inequality we can conclude that the sequence converges in $X_2$ (to $x$, of course.). $$\|x_n - x\|_2 \le c_2 \|x_n - x\|_1 \to 0.$$
This shows that $X_2$ is complete.
Swapping the roles of $X_1$ and $X_2$ we get the desired result.


Addendum#2: Let's consider the statement "Let $(X,\|\cdot\|_1),(X,\|\cdot\|_2)$ be homeomorphic, then there are $\epsilon$ and $\delta$ such that $B^1_\delta(0) \subset B_1^2(0)$ and $B_\epsilon^2(0)\subset B_1^1(0)$."
I can't prove it, and I actually think it is false. Here's why:


*

*there are $\epsilon$ and $\delta$ such that $B^1_\delta(0) \subset B_1^2(0)$ and $B_\epsilon^2(0)\subset B_1^1(0),$

*$\text{Id} \colon X_1 \to X_2$ is a homeomorphism.


$(1)$ and $(2)$ are equivalent, indeed if $\text{Id} \colon X_1 \to X_2$ is a homeomorphism then $\text{Id}^{-1}(B^2_1(0))$ is open in $X_1$ and contains $0$, hence there is $\delta$ such that $B^1_{\delta}(0) \subset \text{Id}^{-1}(B^2_1(0)) = B^2_1(0)$. (the other inclusion is proved similarly)
On the other hand, let $V \subset X_2$ be open and let $0 \in V$. By assumption there is $\eta$ such that $B^2_{\eta}(0) \subset V$. By $(1)$, $B^1_{\eta\delta}(0) \subset B^2_{\eta}(0) = \text{Id}^{-1}(B^2_{\eta}(0)) \subset \text{Id}^{-1}(V)$. In other words, $\text{Id}$ is continuous at $0$. Of course $(1)$ is also equivalent to 


*there are $\epsilon$ and $\delta$ such that $B^1_\delta(x) \subset B_1^2(x)$ and $B_\epsilon^2(x)\subset B_1^1(x)$


just by adding $x$ on both sides. This shows that $\text{Id}$ is continuous. Using the other inclusion we get that $\text{Id}^{-1}$ is continuous as well hence proving that $\text{Id}$ is a homeomorphism.
This shows that if you prove the quoted statement, you could infer that the homeomorphism is given by the identity map, which is a stronger statement.
