Found this question
"Suppose $u, v ∈ V$ are such that $N (u) ⊆ N (v)$, prove that $G$ is not $k$-critical for any $k$"
Now if $G$ was a complete graph $K_n$ such as $K_5$ where there are 5 vertices with edges connecting all of them. Am I wrong in saying the Neighborhood of any 2 vertices would just be the complete graph again? Which still satisfies it being a subset as they're equal right?
In which wouldn't the graph be $k$-critical since, removing any vertex would reduce it to $k_{n-1}$, eg:$ k_5$ would become $k_4$ where the chromatic number, $k$, is $n$ for a complete graph, eg: $k(g) $goes from 5 to 4, making it $k$-critical.