Homeomorphism in a normed space

Question

Let $X$ a normed space. Prove that

$$f(x) = \frac{x}{1+\Vert x \Vert} $$

is a homeomorphism between $X$ and $U=B(0,1) = \{ x\in X / \Vert x \Vert < 1 \}$.

$\textbf{My attempt:}$ Consider $g(x)=\dfrac{x}{1-\Vert x \Vert}$, then $f \circ g =$id$|_{U}$ and $g \circ f =$id$|_{X}$ so $g=f^{-1}$.

Consider $\alpha: X \rightarrow \mathbb{R}$, $\alpha(x)=1+\Vert x \Vert$. How $\alpha(x) \neq 0$ then $\dfrac{1}{\alpha}$ is continuous.

Also, $h_1 : X \rightarrow X$, $h_1(x)=x$ is continuous. So $f=\alpha h_1 : X \rightarrow U$ is continuous.

Analogously consider $\beta: U \rightarrow \mathbb{R}$, $\beta(x)=1- \Vert x\Vert$. How $\beta(x) \neq 0$ then $\dfrac{1}{\beta}$ is continuous.

Also, $h_2 : U \rightarrow X$, $h_2(x)=x$ is continuous. So $g=\beta h_2: U \rightarrow X$ is continuous.

It's correct?

Answer 1

I would prove that $f$ is bounded and injective so it is continous. Note that $f(X) =B(0,1)$, so it it enough to prove that $f^{-1}$ is bounded

Answer 2

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct. However, you should explicity verify that $f$ and $g$ are inverse to each other (which is admittedly fairly trivial).