Isometry of a metric space with proper subset

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 1

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 $\langle X_A,\rho_A\rangle$.