The equality $\chi(G-v)=\chi(G)$

Question

Let $G$ be a graph and $\deg(v)<\chi(G)-1$. By $\deg(v)$ and $\chi(G)$, I mean the degree of vertex $v$ and chromatic number of the graph $G$, respectively. I want to show that $\chi(G-v)=\chi(G)$.

Answer 1

This is true vacuously for graphs with one vertex. Suppose then that $G$ has $N$ vertices and assume that $\chi(G-v)=k<\chi(G)=n$. Without loss of generality assume $k=n-1$. Color $G-v$ with $n-1$ colors. Then $v$ is adjacent to vertices with at most $n-2$ different colors since it has degree at most $n-2$. Then we can color $v$ with the $n-1$st color, obtaining a proper coloring of $G$ with fewer than $\chi(G)$ colors, which is a contradiction. Thus $\chi(G-v)=\chi(G)$.

Answer 2

The inequality $\chi(G-v)\le\chi(G)$ is clear, so you want to show that $$\deg(v)+1\lt\chi(G)\implies\chi(G)\le\chi(G-v).$$ This follows immediately from the obvious inequality $$\chi(G)\le\max\{\chi(G-v),\ \deg(v)+1\}$$ and a simple arithmetical fact: $$z\lt x\le\max\{y,z\}\implies x\le y.$$