Let \(X\) be a graph in which each vertex is an endpoint of only finitely many edges. Show that the weak topology on \(X\) is a metric topology.
Solution
Give every edge of \(X\) length \(1\). For points \(x,y\) in the same connected component, let \[ \rho(x,y) \] be the infimum of the lengths of edgewise paths joining them, and put \(\rho(x,y)=\infty\) when they lie in different components. Define \[ d(x,y)=\min\{\rho(x,y),1\}. \]
1. The function \(d\) is a metric on \(X\).
Symmetry and positive definiteness are immediate from the edge-length construction. For the triangle inequality, the extended path metric \(\rho\) satisfies \[ \rho(x,z)\le \rho(x,y)+\rho(y,z). \] For nonnegative extended real numbers, \[ \min\{a+b,1\}\le \min\{a,1\}+\min\{b,1\}, \] so truncating at \(1\) preserves the triangle inequality. Distinct points on the same finite edge have positive path distance, and points in different components have distance \(1\).
2. The metric topology is contained in the weak topology.
On the closure of each edge, the restriction of \(d\) gives the usual interval topology locally, since for distances less than \(1\) the metric is just edgewise path length. Therefore every metric-open set has open intersection with every closed edge. It also has open intersection with the \(0\)-skeleton. By the definition of the weak topology on a graph, it is weakly open.
3. If \(U\) is weakly open and \(x\) lies in the interior of an edge, then some \(d\)-ball about \(x\) lies in \(U\).
The intersection of \(U\) with that closed edge is open in the interval topology. Choose \(\epsilon>0\) smaller than both the distance from \(x\) to the two endpoints and the interval radius contained in \(U\). Then \[ B_d(x,\epsilon) \] stays inside that edge and lies in \(U\).
4. If \(U\) is weakly open and \(v\) is a vertex of \(X\) lying in \(U\), then some \(d\)-ball about \(v\) lies in \(U\).
Only finitely many edges \[ e_1,\dots,e_r \] are incident to \(v\). For each \(e_i\), weak openness gives a number \(\epsilon_i>0\) such that the initial segment of \(e_i\) of length \(\epsilon_i\) from \(v\) lies in \(U\). If \(r>0\), set \[ \epsilon=\min\{1/2,\epsilon_1,\dots,\epsilon_r\}>0. \] Any point at \(d\)-distance less than \(\epsilon\) from \(v\) must lie on one of these incident edge segments, hence lies in \(U\). If \(v\) is isolated, then \[ B_d(v,1/2)=\{v\}\subseteq U. \] Thus \(B_d(v,\epsilon)\subseteq U\) in all cases.
5. Hence the weak topology equals the topology induced by \(d\).
By <1>3–<1>4 every weakly open set is metric open, while 2 gives the reverse inclusion. Therefore the weak topology is metrizable.