Proving equivalent completeness of two sets Let $X$ and $Y$ be normed space, and $A: X\rightarrow Y$ bounded bijection such that $A^{-1}$ is also bounded. I want to prove that
$$X \; \text{complete} \Leftrightarrow Y \; \text{complete}$$
My work so far
First - I tried to apply somehow Hahn - Banach theorem, but I had no idea how it can be applied. Then my plan was to do this strictly by definition. So:
$\Leftarrow$
Let's assume that $X$ is not complete $\Rightarrow \exists{x_n  \subset X \; \text{Cauchy sequence}}: x_n \rightarrow x \notin X$.
But then if we use our operator $A$ and use fact that its continuous we have $A(x_n) \rightarrow A(x) \notin Y$.
And here I thought that it's the end of this implication proof, because I found Cauchy Sequence that disconverge in Y, however $A(x_n)$ doesn't have to be Cauchy sequence, even if $A$ is bijection.
Could you please give me a  hand, what would be the reasonable approach to this problem?

Well, there's no need for Hahn-Banach or anything like that. The proof is elementary:
Assume that $X$ is complete. We will show that $Y$ is complete. Let $(y_n)\subset Y$ be a Cauchy sequence. Since $A$ is bijective, there exist unique $(x_n)\subset X$ such that $A(x_n)=y_n$ for all $n$.
We show that, since $(y_n)$ is Cauchy in $Y$, $(x_n)$ also has to be Cauchy in $X$, using the definition. Let $\varepsilon>0$. Consider the quantity $\frac{\varepsilon}{\|A^{-1}\|}$ which is positive and apply the definition of Cauchy sequence to this quantity: there exists $n_0\in\mathbb{N}$ such that $\|y_n-y_m\|<\frac{\varepsilon}{\|A^{-1}\|}$ for all $n,m\geq n_0$. Now for $n,m\geq n_0$ we have that
$$\|x_n-x_m\|=\|A^{-1}y_n-A^{-1}y_m\|=\|A^{-1}(y_n-y_m)\|\leq\|A^{-1}\|\cdot\|y_n-y_m\|<\varepsilon. $$
We started by a $\varepsilon>0$ and found $n_0$ such that $\|x_n-x_m\|<\varepsilon$ for all $n,m\geq n_0$, i.e. $(x_n)$ is Cauchy.
Now since $X$ is complete, $(x_n)$ is convergent to some $x\in X$, so $\|x_n-x\|\to0$. We now claim that $y_n\to A(x)$. Indeed,
$$\|y_n-A(x)\|=\|A(x_n)-A(x)\|=\|A(x_n-x)\|\leq\|A\|\cdot\|x_n-x\|\to0$$
We started by an arbitrary Cauchy sequence in $Y$ and proved that it is convergent. Thus $Y$ is complete.
A symmetric argument by replacing $X$ with $Y$ and $A$ with $A^{-1}$ and $A^{-1}$ with $A$ will also show the converse: if $Y$ is complete, then so is $X$. Is everything clear?
