example of not surjective isometry on a bounded not closed set.

Question

Is there an example of a bounded not closed set $A\subset \mathbb R^n$ for some $n\in \mathbb N$ and an isometry $f:A\rightarrow A$
($\forall x,y\in A\space;\space d(x,y)=d(f(x),f(y))$ where $d$ is the euclidean metric) such that $f$ isn't surjective?

Answer 1

Yes. Let $n=2$, so $\mathbb R^n\cong \mathbb C$ and let $f$ be multiplication with $c:=\frac 45+\frac35 i$. Let $A=\{\,c^n\mid n\in\mathbb N_0\,\}$. (The key point is that $c^n\ne 1$ for all $n$, so many different choices of $c$ with $|c|= 1$ work as well)

Answer 2

The idea here is that you need some way to "march off to infinity" without actually going to infinity. We can't do this with $n=1$ (no bounded set can map to itself via an ambient isometry), so let's turn to $n=2$ and irrational rotations. Let $$A = \{e^{i\alpha n}\ | n\in\mathbb{N},\alpha/2\pi\notin\mathbb{Q}\}.$$

Multiplication by $e^{i\alpha}$ is an isometry carrying $A$ to itself which is not surjective, since $1$ is in $A$ but is not in the image because the rotation is irrational.

You can find a more exotic example in the proof of the Banach-Tarski paradox, which as I recall involves constructing an isometry of the sphere to a proper subset of itself.