Kunen exercise III.8.21 Let $f: \omega_1\to \mathbb{R}$ be one-one. Let $g:[\omega_1]^2\to 2$ be such that for any $\alpha<\beta<\omega_1$, $g(\{\alpha, \beta\})$ is $0$ when $f(\alpha)<f(\beta)$, and $1$ otherwise. $g$ yields the following partial orders on finite subsets of $\omega_1$.
$\mathbb{P_\mu} = \{p\in [\omega_1]^{<\omega}: \forall \alpha, \beta\in p(\alpha\not = \beta \to g(\{\alpha, \beta\}) = \mu)\}$
The exercise is to show that neither $\mathbb{P_0}$ nor $\mathbb{P_1}$ are ccc. The order is $p\leq q$ iff $p\supseteq q$.
I know I'm missing something very simple, but I've spent an unhealthy amount of time on this now, and I need to move on!

Here's a forcing argument (not how Kunen wants you to do it since forcing is covered in chapter 4): First, by throwing away all the rational intervals in which $f[\omega_1]$ is countable, we can choose $X \subseteq \omega_1$ such that $\omega_1 \setminus X$ is countable and for every $x \in f[X]$, $|[x, \infty) \cap f[X]| = \omega_1$. Let $Q_0 = \{p \in P_0 : p \subseteq X\}$. Note that for any two conditions $p, q \in Q_0$, $p, q$ are compatible in $Q_0$ iff they are compatible in $P_0$ (as witnessed by their union). Hence it is enough to show that $Q_0$ is not ccc. But this is clear since $Q_0$ adds an increasing $\omega_1$-sequence of reals and hence an injection from $\omega_1$ to rationals so that $\omega_1$ is collapsed. The argument for $P_1$ is identical.
