Disjoint union of proper metric spaces

Question

A proper metric space is one in which closed balls are compact. Given a countably infinite collection of proper metric spaces, can their disjoint union be equipped with a proper metric?

Answer 1

Yes. Choose a family of points $x_m\in X_m$ and call $d:\coprod\limits_{n\in\Bbb N} X_n\times\coprod\limits_{n\in\Bbb N} X_n\to[0,\infty)$ $$d((a,m),(b,n))=\begin{cases} \lvert n-m\rvert+d_m(a,x_m)+d_n(b,x_n)&\text{if } n\ne m\\ d_m(a,b)&\text{if }n=m\end{cases}$$

The idea corresponds to seeing $\coprod_{m\in\Bbb N}X_m$ as a subspace of a larger space where each $X_m$ is connected to $X_{m+1}$ by a bridge of length $1$ from $x_m$ to $x_{m+1}$, the crossing of which is mandatory to change island.

For triangle inequality, notice that:

• $d((x,m),(z,m))+d((y,m),(z,m))=d_m(x,z)+d_m(y,z)\ge d_m(x,y)=d((x,m),(y,m))$

• if $k\ne m$, then \begin{align}&d((x,m),(z,k))+d((y,m),(z,k))=d_m(x,x_m)+2\lvert m-k\rvert+d_m(y,x_m)+2d_k(z,x_k)\ge\\&\ge d_m(x,x_m)+d_m(y,x_m)\ge d_m(x,y)=d((x,m),(y,m))\end{align}

• if $n\ne m$, $n\ne k$ and $k\ne m$, then \begin{align}&d((x,m),(z,k))+d((z,k),(y,n))=\\&=\lvert k-m\rvert+\lvert n-k\lvert +d_m(x,x_m)+2d_k(z,x_k)+d_n(y,x_n)\ge\\ &\ge \lvert n-m\rvert+d_m(x,x_m)+d_n(y,x_n)=d(x,y)\end{align}

• if $n\ne m$, then \begin{align}&d((x,m),(z,m))+d((z,m),(y,n))=\\&= d_m(x,z)+\lvert n-m\rvert +d_n(y,x_n)+d_m(z,x_m)\ge\\& \ge \lvert n-m\rvert+d_m(x,x_m)+d_n(y,x_n)=d((x,m),(y,n))\end{align}

This exhausts all the cases.

This metric induces the disjoint union topology. In order to prove that, we just need to check that the tautological maps $\iota_m:(X_m,d_m)\to \left(\coprod\limits_{n\in\Bbb N} X_n,d\right)$ defined by $\iota_m(x)\mapsto (x,m)$ are open and continuous. They are continuous because, by definition, they are isometries. Moreover, for all $m$, for all $x\in X_m$ and for all $0<\varepsilon<1$, it holds $\iota_m\left[B_{d_m}(x,\varepsilon)\right]=B_d((x,m),\varepsilon)$.

As a result, every $\iota_m$ is a homeomorphism onto a clopen subset. Now, every closed ball $E_d((x,m),r)$ is compact, because $d((x,n),(y,m))\ge \lvert n-m\rvert$ for all points and $E_d((x,m),r)\cap X_n\subseteq \iota_n\left[E_{d_n}(x_n,r)\right]$ for all $n\ne m$. Thus, $$E_d((x,m),r)=\bigcup_{k=\max(\lceil m-r\rceil,0)}^{\lfloor m+r\rfloor}\ E_d((x,m),r)\cap X_k\subseteq\iota_m\left[E_{d_m}(x,r)\right]\cup\bigcup_{k=\max(\lceil m-r\rceil,0)}^{\lfloor m+r\rfloor}\iota_k\left[E_{d_k}(x_k,r)\right]$$

The rightmost set is finite union of compact sets, thus compact. $E_d((x,m),r)$ is closed in a compact set, therefore compact.