Question about a Ramsey graph.

[cragar]

the Ramsey number of $R(\omega,\omega)=\omega$
but then $R(\omega+1,\omega)=\omega_1$
My question is on the second one can we do a counter example to show that it can't
be any countable ordinal.
depending on how I count the natural numbers I can get any countable ordinal I want.
If we assume that $R(\omega+1,\omega)$ was equal to some countable ordinal
then I could just color $\omega$ edges with blue for example and i would stay
under the $\omega+1$ limit . And the other color we would just use a finite number of them.
I guess i don't really understand why order matters for an infinite Ramsey graph.
It doesn't seem like it matters in the finite case.

 

[Valkarie]

The Ramsey number of R(\omega,\omega) is indeed equal to \omega, however this does not necessarily mean that R(\omega+1,\omega) is equal to \omega_1. This is because the Ramsey number of R(\omega+1,\omega) can depend on the specific ordering of the edges in the graph. For example, if we order the edges of the graph so that the two sets of \omega+1 edges are adjacent, then the Ramsey number of R(\omega+1,\omega) will be \omega_1. However, if we order the edges so that the two sets of \omega+1 edges are separated by some other edges, then the Ramsey number of R(\omega+1,\omega) could be lower than \omega_1. Therefore, it is not possible to provide a counter-example to show that R(\omega+1,\omega) cannot be any countable ordinal, as the actual value can depend on the ordering of the edges.