An arrangement of $s$ lines are drawn in the plane so that no three lines intersect at a common point and no two lines are parallel. Now circumscribe this arrangement by a circle so that all intersections are in the interior of the circle. Denote the circumscribed line arrangement by $\mathcal{H}_{s}$. Is this arrangement $3$ $vertex-connected$ ?
Here was my attack. Let $x$ be any point on the boundary and suppose that $x$ is adjacent to the boundary points $a$ and $b$ and an interior point say $c$. Now $deg(x)=3$. Removal of $a$, $b$ and $c$ disconnects $x$ from $H_{s}$? In particular $(a,b,c)$ is a vertex cut?
I am not certain I understand vertex-connectivity correctly and I might be confusing it with edge-connectivity?
As you guess the vertex-connectivity of the above graph is 3.
The vertex connectivity of graph $G$ denoted by $\kappa(G)$ is the minimum number of vertex whose removal yields to disconnected graph. In analogical way, the edge connectivity of graph $G$ denoted by $\lambda (G)$ is the minimum number of edges whose removal yields to disconnected graph.They are related in the following inquality: $$\kappa(G) \leq \lambda (G) \leq \delta(G)$$ where $\delta(G)$ is the minimum degree of $G$
Example:
-Any tree has vertex connectivity equal to 1.
-The complete graph on $n$ vertices has edge-connectivity equal to $n-1$.