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The neighbourhood of a vertex $v$ is the set of all neighbours of $v$ (or, equivalently, the set of vertices adjacent to $v$), and is denoted by $N(v)$. Prove that if $u, v$ are vertices in a critical graph $G$, then $N(u)⊈ N(v)$.

A hint was provided: The proof deals with two cases of $u$ and $v$ are adjacent and not adjacent separately. One case is trivial.

Any help will be appreciated.

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    @rschwieb I have changed the title as you suggested, any help with the question? – 2017-01-16
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    @user407151 Nope: not until I find a bit of time to look up a few definitions. But in any case I think you're a lot better off with this title! :) – 2017-01-16
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    @bof thank you 'not a subset' is what i meant. – 2017-01-16
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    I don't understand your two cases. If $N(u)\subseteq N(v)$ then $u$ and $v$ can't be adjacent; $v\notin N(u)$ because $v\notin N(v).$ (I assume your graphs are loopless since the problem is about chromatic number.) – 2017-01-16
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    A graph G is critical if χ(H) < χ(G) for every proper subgraph H of G. We also say a critical graph G with χ(G) = k is k-critical. That's the definition i was given in my notes. – 2017-01-16
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    can you show this futher how it explains that N(u)⊈N(v) as im still finding it confusing @bof. thank you for your help. – 2017-01-16
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    Im not sure @bof – 2017-01-16
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    Thank you for your help it was much appreciated @bof. If you have another suggestions regarding this proof please do let me know. – 2017-01-16
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    I was also given a theorem which states 'If G is K-critical, then δ(G)≥K-1.' which i was told would be useful to solve this. Is this theorem similar to what you have suggested? @bof – 2017-01-16
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    If $\chi(G)=k$ and some vertex $u$ has $\lt k-1$ neighbors, then deleting $u$ can't lower the chromatic number; if you could color the rest of the vertices with $k-1$ colors, there would be no problem coloring $u$. So in a $k$-critical graph every vertex has degree at least $k-1.$ I don't see how that would help to prove the thing you asked about in your question. – 2017-01-16
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    Ive noticed that the proof for that theorem is similar to the proof you gave for this question hence why it was probably given as a hint. Thank you again. @bof – 2017-01-16

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Let $G$ be a critical graph. Assume for a contradiction that there are vertices $u,v\in V(G)$ with $u\ne v$ and $N(u)\subseteq N(v).$ Let $h=\chi(G-u).$ Consider a proper coloring of $G-u$ with $h$ colors. Since $N(u)\subseteq N(v),$ this coloring can be extended to a proper coloring of $G$ with $h$ colors, by giving $u$ the same color as $v.$ This shows that $\chi(G)=\chi(G-u),$ contradicting the assumed criticality of $G.$