Graph: question on planar graph. I have a lemma that say:
Let $G$ be a planar graph whose exterior face is bounded by a cycle $u_1,...,u_k$. Then there is a vertex $u_i$ ($i\neq 1,k$) not adjacent to any $u_j$ other than $u_{i-1}$ and $u_{i+1}$.
But for example $K_4$ is planar and this condition doesn't look satisfy. So I don't really understand this lemma.

 A: As it is written, the graph $K_4$ does satisfy the lemma.
Below I have a planar representation of $K_4$. The exterior face is bounded by the cycle $u_1$, $u_2$, $u_3$. Here we have $k=3$, and then a vertex $u_i$, such that $i\neq 1, k$, has to be $u_2$. Note how this vertex is not adjacent to any $u_i$ other than $u_1$ and $u_3$.
The key point is of course that the fourth (unnamed) vertex of $K_4$ is not part of the bounding cycle of the exterior face.

EDIT: Here's the usual planar representation of $K_4$. In this representation, it is easier to see that the fourth vertex does not take part in the cycle bounding the exterior face. The lemma concerns the vertices $u_1,u_2,\ldots,u_k$, but these consist only of the vertices of your graph $G$ that bound the exterior face. Notice how the center vertex is not part of the bounding cycle, and thus the lemma does not concern this vertex.

Also, here's your other example (you had the vertex $v_3$ in two places, so I changed one of them to $v_2$). Notice how the exterior face is bounded by the vertices $v_3$, $v_4$, and $v_5$. Thus, if you let $u_1=v_3$, $u_2=v_4$, and $u_3=v_5$, then the lemma says something about the cycle $u_1,u_2,u_3$, but nothing about the remaining vertices.

