I've been reading through a book called graph homomorphism and this is an exercise I've been trying to prove.
Here's my work so far
Induction on number of vertices
Basis : this clearly holds
Induction :Let G have n+1 vertices. let X and Y be two sets in G compliment that have no edges between them. (since G compliment is disconnected these sets exist)
Consider the subgraph induced by G[X]. By induction hypothesis, the chromatic number = circular number of G[X]
Similarly for G[Y].
Since X, Y have no edges between them in G compliment, in G every vertex in X must be adjacent to every vertex in Y.
Therefore we can show the chromatic number of G = chromatic number of G[X] + G[Y] as every colour used in set X cannot be used on the vertices of set Y and vica versa. Now we just have to show that this is equal to the circular colouring.
Here's where I can't get anywhere so any help is appreciated.
Other approaches I took were to Induct on the chromatic number and then remove one of the partitions and see how it goes and a proof by contradiction using the circular arc definition.