By Ramsey's theorem we know that:
$\forall k \in \mathbb N : \exists N \in \mathbb N$ that an arbitrary graph $G$ on a set of vertices $\{1,2,...,N\}$ contains $k$ vertices, which represent either a complete subgraph or an independent set.
Now I need to decide, whether it's true or not that:
a, - amongst these $k$ vertices are vertex number $1$
b, - all these $k$ vertices are vertices with even numbers
I'm having difficulties proving or finding a counterexample for these and I'm not even sure that I understand the problem completely (of course, if the first one is true, then the second is false). Any hint/thought is appreciated.
Assuming $k>2$. No matter how big $N$, if you take your graph $G$ with $N$ vertices to have only the edges $\{1,i\}$ for $i=2,\ldots,N$, then no complete subgraph of order $k$ will contain the vertex 1 and no independent set of order $k$ will contain the vertex 1, so the first statement is false.
The second statement is true: if you have found an $N_k$ for the 'regular' theorem, simply use $2N_k$ for the theorem restricted to even numbers: the subgraph induced by the vertices of even numbers is now an arbitrary graph with $N_k$ vertices, so you can apply the regular theorem.