# How to know that given a simple graph $G$ that a vertex $v$ if connected, then $G - v$ is also connected.

I'm trying to understand the answer to this question:

If $$G$$ is simple with diameter two and maximum degree $$|V(G)| - 2$$, then $$|E(G)| \geq 2|V(G)| - 4$$

The person that answered claimed that if $$deg(v)=n-2$$ then $$G-v$$ is connected, but I just don't understand that part (highlighted below):

"It is easy to see that $$G−v$$ is a connected graph. (There is a vertex $$u$$ of $$G−v$$ which is not adjacent to $$v$$; every other vertex of $$G−v$$ is connected to $$u$$ by a path in $$G$$ of length at most $$2$$; since there is no edge $$uv$$, that path must be contained in $$G−v$$.)..."

• the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance. Commented Jan 17, 2019 at 13:25

Let $$G$$ with $$diam(G)=2$$ and max degree $$\Delta(G)=n-2$$. Then let $$v$$ a vertex such that $$d(v)=n-2$$.
There exist a unique $$u$$ such that $$v$$ and $$u$$ are not connected, $$v \nsim u$$.
Now, for all $$w\in V(G-\{v\})$$, we know that in $$G$$, the distance between $$w$$ and $$u$$ is at most 2. But as $$u$$ is not connected to $$v$$, the 2-edges-path from $$w$$ and $$u$$ cannot pass through $$v$$. Therefore the path from $$u$$ to $$w$$ exists also in $$G-v$$. This is true for any $$w\neq v$$, therefore $$G-v$$ is connected.
Edit Even more : you don't need $$d(v)=n-2$$, you only need $$d(v)\leq n-2$$ to get at least one vertice $$u$$ satisfying the condition. Therefore as long as the maximum degree $$\Delta(G)$$ is at most $$n-2$$ (which is the case here), then for any $$v$$ , $$G-v$$ is connected, and therefore $$G$$ is 2-connected.