The Herschel graph is an undirected graph with 11 vertices and 18 edges. It is the smallest non-Hamiltonian polyhedral graph, meaning it cannot be embedded on a sphere without crossing edges and does not contain a Hamiltonian cycle. [1].

The condition states that if $S$ is a set of vertices of a Hamiltonian graph $G$, then $c(G-S) ≤ |S|$, where $c(G-S)$ is the number of connected components in the graph $G$ after removing the vertices in set $S$.

Apply the condition to the Herschel graph
To show that the Herschel graph is non-Hamiltonian, we need to find a set of vertices S such that $c(G-S) > |S|$.

Find the set of vertices $S$.
In the Herschel graph, we can choose the set $S = {v_1, v_2, v_3}$ where $v_1$, $v_2$, and $v_3$ are three non-adjacent vertices. After removing these vertices, the remaining graph will have 8 vertices.

Calculate $c(G-S)$ and $|S|$.
After removing the vertices in set S, the Herschel graph will be divided into 4 connected components. So, $c(G-S) = 4$. The size of the set S is $|S| = 3$.

Compare $c(G-S)$ and $|S|$.
Since $c(G-S) = 4$ and $|S| = 3$, we have $c(G-S) > |S|$. This means that the Herschel graph does not satisfy the given condition for a Hamiltonian graph [1].