# Question related to complete vertex, interior vertex and boundary vertex of a graph

All the definition and results which I am using here has been taken from the Garry Chartrand book $$\textbf{The Introduction of Graph Theory}$$ and the paper $$\textbf{On the Commuting Graph of Dihedral Group}$$. Here I am denoting the set N$$(v)$$ is the collection of all vertices which are adjacent to $$v$$ in a simple graph $$\Gamma$$ and the symbol $$u \sim v$$ means $$u$$ and $$v$$ are adjacent.

• A vertex $$v$$ in a graph $$\Gamma$$ is a $$\textbf{boundary vertex}$$ of a vertex $$u$$ if $$d(u, w) \leq d(u, v)$$ for $$w \in$$N$$(v)$$, while a vertex $$v$$ is a boundary vertex of a graph $$\Gamma$$ if $$v$$ is a boundary vertex of some vertex of $$\Gamma$$.

• A vertex $$v$$ is said to be a $$\textbf{complete vertex}$$ if the subgraph induced by the neighbors of $$v$$ is complete.

• A vertex $$v$$ is said to be an $$\textbf{interior vertex}$$ of a graph $$\Gamma$$ if for each $$u \ne v$$, there exists a vertex $$w$$ and a path $$u-w$$ such that $$v$$ lies in that path at the same distance from both $$u$$ and $$w$$. A subgraph induced by the interior vertices of $$\Gamma$$ is called \emph{interior} of $$\Gamma$$ and it is denoted by $$Int(\Gamma)$$.

The following results are on the page of 337 and 339 of the above book.

Let $$\Gamma$$ be a connected graph and $$v \in V(\Gamma)$$. Then $$v$$ is a complete vertex of $$\Gamma$$ if and only if $$v$$ is a boundary vertex of $$x$$ for all $$x \in V(\Gamma) \setminus \{v\}$$.

Let $$\Gamma$$ be a connected graph and $$v \in V(\Gamma)$$. Then $$v$$ is a boundary vertex of $$\Gamma$$ if and only if $$v$$ is not an interior vertex of $$\Gamma$$.

If $$\Gamma$$ is a complete graph of size $$n$$ and $$v \in V(\Gamma)$$, then by the definition of complete vertex, $$v$$ is a complete vertex. By using the above two results, $$v$$ is not an interior vertex.

when I am applying the definition of an interior vertex, for each $$u \ne v \in V(\Gamma)$$ and chose $$w \in V(\Gamma) \setminus \{u, v\}$$, we have a path $$u \sim v \sim w$$. Thus, $$v$$ is an interior vertex.

I am a little bit confused so please help me where I am doing wrong. I would be thankful for your kind help.

I suspect the issue is with the definition of interior you are using. This should say that $$v$$ is an interior vertex of $$\Gamma$$ if for each $$u\neq v$$, there exists a vertex $$w$$ and a $$u-w$$ geodesic containing $$v$$.
With this definition, the path $$u\sim v\sim w$$ in $$K_n$$ does not cause any problems, because it is not a geodesic (the single edge $$u\sim w$$ is a shorter path).