# Two blocks in a graph share at most one vertex

I'm reading Introduction to Graph Theory by Douglas West, and one of the propositions is

Two blocks in a graph share at most one vertex.

A block of a graph $$G$$ is a maximal connected subgraph of $$G$$ that has no cut-vertex.

The proof is as follows:

We use contradiction. Suppose that blocks $$B_1$$, $$B_2$$ have at least two common vertices. We show that $$B_1 \cup B_2$$ is a connected subgraph with no cut vertex, which contradicts the maximality of $$B_1$$ and $$B_2$$. When we delete one vertex from $$B_i$$, what remains is connected. Hence we retain a path in $$B_i$$ from every vertex that remains to every vertex of $$V(B_1) \cap V(B_2)$$ that remains. Since the blocks have at least two common vertices, deleting a single vertex leaves a vertex in the intersection. We retain paths from all vertices to that vertex, so $$B_1 \cup B_2$$ cannot be disconnected by deleting one vertex.

I'm confused as to where $$B_i$$ comes from in this proof. Is it different from $$B_1 \cup B_2$$? If so, what is it? That paragraph is the first time $$B_i$$ showed up in the section.

• Bi is just shorthand for B1 or B2. Sep 28, 2018 at 22:23

If we can find two distinct vertices in $$B_1 \cap B_2$$, then the idea is that removing a vertex in the intersection maintains at least another vertex to connect $$B_1$$ with $$B_2$$, thus 2-connectivity of $$B_1 \cup B_2$$ (which is a larger 2-connected subgraph).