Dushnik-Miller Proof I am working through the Dushnik-Miller Theorem in Jech's Set Theory and have a question to one part of the proof. For the one's who don't have the book at their fingertips here Theorem 9.7:
For every infinite cardinal $\kappa$ holds $\kappa\to (\kappa, \omega)^2$.

For the proof he choses $\{A,B\}$ to be the partition of $[\kappa]^2$ and for every $x\in\kappa$ $B_x := \{y\in\kappa:x<y \textrm{ and } \{x,y\}\in B \}$.

He differs two cases.
The first is: for all $X\subseteq \kappa$ with cardinality $\kappa$ exists $x\in X$ such that $|B_x \cap X|= \kappa$ 
The proof of this part is clear to me.

The second is: There exists a $S\subseteq \kappa$ with cardinality $\kappa$ such that $|B_x \cap S|<\kappa$ for all $x\in S$.
In this part he differs between $\kappa$ regular and singular.
Here is my question. I don't really see why we do this distinction...

I hope someone can explain me with some more details this part of the proof!

Thank you, and all the best

Luca!
 A: I'm a couple thousand miles away from my books, so the following may not match what Jech does, but it's how I think of the proof.  The reason for distinguishing the regular and singular cases is that the regular case is easy and the singular case isn't (or at least it's less easy). Here's the allegedly easy argument in the regular case. I'll choose, by induction on ordinals $\alpha<\kappa$, elements $s_\alpha$ of $S$ (where $S$ is as in your case hypothesis) such that each two-element set of the form $\{s_\alpha,s_\beta\}$ is in $A$. The inductive process is easy because it can be described as "just don't do anything stupid."  In more detail: Pick $s_0\in S$ arbitrarily. Now since we want $\{s_0,s_\beta\}$ to be in $A$ for all $\beta>0$, it would be stupid to even consider picking any future $s_\beta$ from $B_{s_0}$. So discard all those stupid elements from $S$.  What's left, $S-B_{s_0}$, still has size $\kappa$ (as $|B_{s_0}|<\kappa$); in particular it's nonempty, so we can pick an arbitrary $s_1$ from it.  Now, to avoid stupidity, we have to discard all elements of $B_{s_1}$, but there still remain $\kappa$ elements, and we can pick one to serve as $s_2$.  Continue in this manner, alternately picking $s$'s and discarding the associated sets $B_s$,  for $\kappa$ steps, and you'll be done, provided you never run out of elements to pick, i.e., provided you never find yourself having discarded all of $S$.  Well, at any stage $\alpha$ of the process, you've discarded only $|\alpha|<\kappa$ sets, each of size $<\kappa$. By regularity of $\kappa$, that's $<\kappa$ elements altogether, so some elements of $S$ have survived and you can pick $s_\alpha$.  
If $\kappa$ were singular, you might have discarded all the elements of $S$ already at stage cf$(\alpha)$.  So the singular case requires a more sophisticated argument.
