Example 3

With the notation of Example 2, suppose now that the $A_{ix}$ are closed and the $h_{x!}$ are homeomorphisms; further suppose that for each index i there is only a finite number of indices x such that $A_{ix} \neq \varnothing$ (i.e. each $X_i$ is "stuck" to only a finite number of $X_x$). Then the relation R is closed. For if F is any closed subset of X, the saturation of F is the union of the sets $h_{x!}(F \cap A_{ix}) \subset A_{ix}$; the assumptions made imply that this family is locally finite, and $h_{x!}(F \cap A_{ix})$ is closed in $A_{ix}$ and therefore in X. The result therefore follows from Proposition 4 of § 1, no. 5.