Cliques and separators
For the six-vertex graph, list \mathrm{bd}(\{1,2\}) and \mathrm{cl}(\{5\}), and state three conditional-independence statements implied by the global Markov property. Is \{2\} a (1,4)-separator?
The edges are 12, 13, 23, 24, 34, 45, 46, 56. The boundary \mathrm{bd}(\{1,2\}) is the set of vertices outside \{1,2\} adjacent to 1 or 2: vertex 3, adjacent to both, and vertex 4, adjacent to 2, so \{3,4\}. The closure is \mathrm{cl}(\{5\})=\{5\}\cup\mathrm{bd}(\{5\})=\{4,5,6\}.
Three global Markov statements: \{4\} separates \{1,2,3\} from \{5,6\}, so X_{\{1,2,3\}}\perp X_{\{5,6\}}\mid X_4; \{2,3\} separates 1 from \{4,5,6\}, so X_1\perp X_{\{4,5,6\}}\mid X_{\{2,3\}}; and the local property at vertex 5, X_5\perp X_{\{1,2,3\}}\mid X_{\{4,6\}}. The pairwise statement X_1\perp X_4\mid X_{\{2,3,5,6\}} is the same idea on a pair. The set \{2\} is not a (1,4)-separator, because the path 1-3-4 avoids 2.