Consider a coloring $c:[\kappa]^2 \to 2$ ($\kappa$ a regular uncountable cardinal, can be assumed to be $\omega_1$ for simplicity) s.t. the following holds:
For every $A \subset [\kappa]^{<\omega} $ of size $\kappa$, consisting of pairwise disjoint sets, and every $i\in\{0,1\}$ there are $a,b\in A$ with $\sup a<\min b$ s.t. $c``[a\times b]=\{i\}$
[If I'm not mistaken, these colorings exist e.g. when $V=L$ and $\kappa$ is not weakly compact]
Now we define a forcing notion $P$ consinting of all finite functions $f:\kappa\to\omega$ satisfying $f(\alpha)=f(\beta) \Rightarrow c(\alpha,\beta)=0$. $f$ is stronger than $g$ if $g\subset f$.
The above condition on $c$ implies $P$ has $\kappa$-cc (so $\kappa$ doesn't collapse). A generic for $P$ gives a partition of $\kappa$ into $\omega$ many sets, each $c$-homogeneous with color $0$.
I am trying to understand what can be said about this partition. A pigeon-hole argument implies there is at least one set in the partition of size $\kappa$. So one can ask:
- Can there be more than one set in the partition of size $\kappa$?
- Can there be only one set in the partition of size $\kappa$?
- Can all the sets in the partition be of size $\kappa$?
- etc.
I am aware that there may be a positive answer to all these questions, i.e. that $P$ doesn't force much about the partition. But if so, how do I prove it? How should one go about in investigating this forcing?
Following Asaf's suggestion, we show that actually it is forced that all sets in the partition are of size $\kappa$.
First, we note that the set $$B=\{a\in[\kappa]^{<\omega} \mid \exists \beta(a)>\sup(a) \ s.t. \forall \beta\geq\beta(a) \exists \alpha\in a,\ c(\alpha,\beta)=1 \}$$ is of size $<\kappa$. Otherwise, we can choose by induction $\{a_\xi \mid \xi<\kappa\}$ by choosing for every $\xi<\kappa$ some $a_\xi\in B$ s.t. $\min(a_\xi)>\sup\{\beta(\alpha_\zeta) \mid \zeta<\xi\}$. So by the definition of the $\beta(a)$ for every $\zeta<\xi<\kappa$ and every $\alpha_\zeta\in a_\zeta$, there is some $\alpha_\xi\in a_\xi$ with $c(\alpha_\zeta,\alpha_\xi)=1$, which contradicts the assumption on $c$ (which states that there are $\zeta<\xi$ with $c``[a_\zeta \times a_\xi]=\{0\}$).
So WLoG, by ignoring some initial segment of $\kappa$, we can assume that for all $a\in[\kappa]^{<\omega}$ there are unboundedly many $\beta$-s s.t. $c(\alpha,\beta)=0$ for every $\alpha \in a$.
This means that for every $n<\omega$ the set $$D_{n,\alpha}=\{f\in P \mid \sup(f^{-1}(n))>\alpha\}$$ is dense in $P$ (for any $f\in P$ apply what we've shown on $dom(f)\in[\kappa]^{<\omega}$ to get an extension $f\cup \{(\beta,n)\}$ with $\beta>\alpha$). So in a generic extension, every set in the partition would be unbounded, and therefore of size $\kappa$.