graph theory
I'm curious whether a simple graph which contains just one vertex is edge k-connected or vertex k-connected.
Edge-k-connectivity: We theoretically can remove any number of edges and it stays connected. The same with vertexes.
Graph satisfies is edge-k-connectivity: IF we remove any k edges -> Graph still satisfies connectivity.
In this case, the left side of implication is never satisfied so it is False. It the left side is False, then the whole implication is True.
EDIT:
Graph -> simple graph
Let $G$ be a graph and let $\kappa(G)$ be the size of any minimum vertex separating set of $G$, $\lambda(G)$ be the minimum size of any edge separating set of $G$, and let $\delta(G)$ be the minimum degree of $G$. It is well known that $$\kappa(G)\le\lambda(G)\le\delta(G).$$ Since $\delta(G)=2$ and $G$ is connected then $\kappa(G)=1\text{ or }2$. Can you show that $\kappa(G)\ne 1$?
When you are drawing graphs, edges should never meet unless they are crossing or share a vertex in common and they should only meet in those locations. The way you have drawn your graph seems to indicate that $\deg(v_5)=3$ when it is really true that $\deg(v_5)=4$. Similarly for $v_7$.
Would separating $A$ into two nodes connected by one edge work? Also, the minimum degree of your graph is $2$ at each of the four corner nodes.
The graph with the largest minimum degree, lowest edge connectivity, and lowest vertex connectivity is formed by connecting two complete graphs with one singular edge like a bridge.
Best Answer
According to conventions that I have seen, edge and vertex connectivity for a single vertex graph is undefined.
If you try to say that the graph is vertex 1-connected or edge 1-connected, you run into an issue.
For example, if a graph is vertex 1-connected, than we can remove one vertex and disconnect the graph. However, if we remove one vertex from the graph, we get the empty graph, which is connected by convention.
Similarly, if we say that the graph is edge 1-connected, then this means that we can disconnect the graph by removing one edge. However, there are no edges to remove, so this doesn't make sense.