We last met the ccc in the context of preserving cofinalities in forcing extensions; but the exercise that I gave the proof for was specifically about forcing. This week, I

ran across a nice little exercise which doesn't explicitly mention forcing at all -- it's pure combinatorics -- but, at least for me, thinking about it using the forcing idea was the key to solving the problem.

**Exercise 1:** Let \( \mathbb{P} \) be a ccc poset, and let \( X = \{ x_i \colon i < \omega_1 \} \subseteq \mathbb{P} \) be a subset. Show that there exists an uncountable subset of \( X \) whose every pair are compatible.

At first glance, this looks completely obvious -- things are either compatible or incompatible, and you can't have more than countably many pairwise incompatible things. The problem, though, is that the compatibility relation need not be transitive -- just because you have an uncountable subset of \(X\) which is not an antichain does

*not* mean that it satisfies the conclusion of the exercise.

What made this problem interesting to me was that I had to

*discard* some of the heuristics that usually serve me well. In particular, a dependable heuristic when dealing with posets is duality: if something is true for all posets, then you should be able to turn your poset upside-down and it should still be true. However, the compatibility relation is not stable under duality! Most forcing posets have a single weakest condition; if you dualize, you get that every two conditions are compatible, which is clearly useless.