In the proof of normality of a linear graph \(X\), why did we assume that every vertex of \(X\) belongs either to \(B\) or to \(C\)?
Solution
Let \(X^0\) be the set of vertices of the linear graph \(X\). As proved just before Lemma 83.1, \(X^0\) is a closed discrete subspace of \(X\), and in fact every subset of \(X^0\) is closed in \(X\).
Given the original disjoint closed sets \(B,C\), let \[ D=X^0-(B\cup C). \] Then \(D\) is closed, so replacing \(B\) by \(B\cup D\) preserves closedness and keeps it disjoint from \(C\). Thus, without loss of generality, every vertex lies in \(B\) or in \(C\).
This extra condition is used twice in the proof. First, if the open sets chosen separately inside two incident edges overlap, their common point must be the common vertex; knowing that every vertex lies in \(B\) or \(C\) shows such a point cannot lie simultaneously in the \(B\)-neighborhood and the \(C\)-neighborhood. Second, when proving the unions of the edgewise neighborhoods are open in the coherent topology, any extra intersection point between neighborhoods from different edges is again a vertex, and membership in \(B\) or \(C\) forces it into the appropriate edgewise neighborhood. Without assigning the unused vertices first, these two arguments would leave the behavior at such vertices uncontrolled.