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?

  • $\begingroup$ Can you prove that such a coloring has to have a homogeneous set of each color with cardinality $\kappa$? $\endgroup$
    – Asaf Karagila
    Apr 3, 2016 at 15:43
  • $\begingroup$ @asaf Do you mean before the forcing? Quite the contrary - it has no homogenous set of size $\kappa$ for any color. This is a private case of the condition, when you take $A$ to be a set of singletons. Then in any such set of size $\kappa$ and any color you choose you can find a pair in the set colored by it. $\endgroup$
    – Ur Ya'ar
    Apr 3, 2016 at 15:48
  • 1
    $\begingroup$ Ah, yes. You're right. I think this is enough to prove that for every $\alpha$ there are unboundedly many $\beta$'s such that $c(\alpha,\beta)=0$. The idea is to show that if this is not the case then you can find a pre-homogeneous set, from which you can actually produce a homogeneous set. And this should give you that every part in the generic partition has size $\kappa$. $\endgroup$
    – Asaf Karagila
    Apr 3, 2016 at 15:56
  • $\begingroup$ Oh Ok, I think I got it. I think initially it's not strictly "for every $\alpha$", but "for all but boundedly many $\alpha$", but then we can WLoG cut off this initial segment to get "for every". This is really what I was missing in my attempts. Do you plan to write it more explicitely as an answer, or should I? $\endgroup$
    – Ur Ya'ar
    Apr 3, 2016 at 17:48
  • $\begingroup$ No, go ahead. I'm too busy to check the details now. $\endgroup$
    – Asaf Karagila
    Apr 3, 2016 at 18:08

1 Answer 1


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 $\kappa$.

  • $\begingroup$ Yeah, that looks okay to me. $\endgroup$
    – Asaf Karagila
    Apr 4, 2016 at 14:43
  • $\begingroup$ Note that there is a loss of generality: You are changing the poset when you ignore the initial segment. So for example, you will not (necessarily) get a partition of $\kappa$ when you force with this poset, as you would with the original poset. $\endgroup$
    – tci
    Apr 4, 2016 at 15:07
  • $\begingroup$ @tci what I mean is that since $\kappa$ minus the initial segment is still of size $\kappa$, by taking a bijection I can replace the original coloring, that satisfies what I want on an end-segment of $\kappa$, with one that satisfies it on all $\kappa$. This is the WLoG. $\endgroup$
    – Ur Ya'ar
    Apr 4, 2016 at 15:42
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    $\begingroup$ Sure. I was only pointing out that this does not technically answer your original question. Indeed, as the elements of $B$ witness, there might be conditions which force that an element of the partition has size less than $\kappa$. If $\bigcup B$ is infinite, we even get that infinitely many elements of the partition will have size less than $\kappa$. $\endgroup$
    – tci
    Apr 4, 2016 at 16:49
  • $\begingroup$ @tci Good point, thanks. $\endgroup$
    – Ur Ya'ar
    Apr 4, 2016 at 19:05

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