Proof
To show that \(\operatorname{Vis}_{\mathbf v}(\mathcal S')\) is a polyhedral complex, we need to show that both the containment property and the intersection property hold. For the containment property, take any face \(G \preceq F\) where \(F \in \operatorname{Vis}_{\mathbf v}(\mathcal S')\). We want to show that \(G \in \operatorname{Vis}_{\mathbf v}(\mathcal S')\). By definition, we know that \(v \not\in T_F(\mathcal{S'})\). Since \(T_G(\mathcal S')\subseteq T_F(\mathcal S')\), then if \(v\in T_G(\mathcal S')\), then it would follow that \(v\in T_F(\mathcal S')\), contradicting the assumption that \(v\notin T_F(\mathcal S')\). Hence \(v\notin T_G(\mathcal S')\), and therefore \(G\in\operatorname{Vis}_{\mathbf v}(\mathcal S')\). Therefore, the containment property holds.
For the intersection property, let \(F,G\in\operatorname{Vis}_{\mathbf v}(\mathcal S')\). Since \(\mathcal S'\) is a polyhedral complex, \(F\cap G\) is a face of both \(F\) and \(G\). In particular, \(F\cap G\preceq F.\) Since we have already shown that \(\operatorname{Vis}_{\mathbf v}(\mathcal S')\) satisfies the containment property, it follows that \(F\cap G\in\operatorname{Vis}_{\mathbf v}(\mathcal S').\) Hence \(F\cap G\) is a face of both \(F\) and \(G\), so the intersection property holds. \(\ \blacksquare\)