Question: In Irving Kaplansky's "Set Theory and Metric Spaces", exercise 17 on page 71 asks for an example of a metric space which is isometric to a proper subset of itself. Any infinite discrete space and any $\ell^p$ space are such spaces, for different reasons. I want some kind of middle ground, that is, some example which is more intriguing than a discrete space and doesn't require knowledge of functional analysis or manifolds. Thanks in advance!

Answer: 
Let $\langle Y,d\rangle$ be any metric space, and fix a point $p\in Y$. Let $A$ be any infinite set, let $P=\{p\}\times A$, and let $X\_A=(Y\times A)/P$, the set resulting from $Y\times A$ by identifying $P$ to a point. By abuse of notation call this point $p$. Define a metric $\rho\_A$ on $X\_A$ as follows:

$$\rho\_A(\langle y,\alpha\rangle,\langle z,\beta\rangle)=\begin{cases}
d(y,z),&\text{if }\alpha=\beta\\
d(y,p)+d(p,z),&\text{if }\alpha\ne\beta\;,
\end{cases}$$

and $\rho\_A(p,\langle y,\alpha\rangle)=d(p,y)$. $\langle X\_A,\rho\_A\rangle$ is a kind of ‘hedgehog’ with $|A|$ spines all radiating from the centre $p$. If $B$ is any subset of $A$ such that $|B|=|A|$, then $\langle X\_B,\rho\_B\rangle$ is isometric with $\langle X\_A,\rho\_A\rangle$.
HINT:
 Let $\theta$ any real number such that $\frac{\theta}{\pi}$ is irrational. The metric space $(e^{i n \theta})\_{n\ge 0}$ is isometric to the subset $(e^{i n \theta})\_{n\ge 1}$ by the map $z \mapsto e^{i\theta} \cdot z$