I am currently working on an assignment for my discrete mathematics lecture. We are currently looking at outerplanar graphs and I am supposed to proof the following.
Let G = (V,E) be a outerplanar graph, that is, a graph which is planar and for which all Vertices lie on the edge bordering the unlimited area. Show:
If G is connected and $|V| \ge 2$, then there is a Vertex with degree $\ge 2$.
Now, the thing that threw me off here is that we do not set $|V| > 2$, but instead only $|V| \ge 2$. Yet, this seems like it would be too little to me, as a graph with two vertices which are connected by a single edge is, if I do not misunderstand anything, planar and its vertices all lie on the same edge. Is there something in the definition that stops this from being a valid outerplanar graph? Because if this is true, then I have constructed a graph for which the condition is not true, as with two vertices I can at most have a max degree = 1.
Thanks for reading!
You are right, the graph $K_2$ is a counter example.
Now, if you assume that $V \geq 3$, and that $G$ is connected, it can be shown that $G$ has a vertex of degree at least 2, no matter if $G$ is outerplanar or not.
This is an immediate consequence of $$2E =\sum_{v \in V} \deg v \\ E \geq V-1$$