Every riemannian manifold has finite distance between points in different components

Given: $M=\mathbb{R}\times \{0,1\}$ with its disjoint-union smooth structure and metric $d{x}^2$ on each line; $p=(0,0)$ and $q=(0,1)$.

Extended riemannian distance on a disconnected manifold: The extended Riemannian distance on arbitrary $M$ is the componentwise Riemannian distance when two points are in the same component, and $+\mathrm{\infty }$ otherwise. Within each component use thm-riemannian-distance-is-a-metric. Components are open, since small coordinate balls are connected. A continuous curve cannot meet two components because its connected interval image is connected, so the cross-component curve family is empty, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. This is an extended metric: if two endpoints are in different components, any third point is in a different component from at least one of them, so the triangle inequality has infinite right side. It is a finite metric precisely when there are no distinct components. Empty and singleton manifolds retain their unique distances.