Antichains and closure properties

There are many different properties that forcings can have, whose consequences are usually well-known. As an example, intuitively, closure properties of forcings yield preservation of cardinals below, and antichain properties yield preservation of cardinals above. But these properties seem mostly to be studied individually, so Stamatis Dimopoulos and I set out to find these folklore results about which combinations of closure properties and antichain properties can consistently hold.

Let’s start off by recalling a few definitions. A forcing notion \mathbb P is {<}\kappa-closed if every chain in \mathbb P of length {<}\kappa has a lower bound in \mathbb  P. We can increase the strength of this by saying that it’s {<}\kappa-directed closed if every directed system of size {<}\kappa has a lower bound in \mathbb P. We can also weaken {<}\kappa closure by considering the following game for \kappa many rounds.

\begin{array}{cccccccccc}\text{I} & 1_{\mathbb P} && p_1 && p_2 && \cdots\\ \text{II} && p_0 && p_3 && \cdots\end{array}

Here p_\alpha\in\mathbb P and p_{\alpha+1}\leq p_\alpha for all \alpha<\kappa. Player I wins iff they can keep on playing throughout all the rounds. We then say \mathbb P is \kappa-strategically closed if player I has a winning strategy in this game. Note that every {<}\kappa-closed \mathbb P is also \kappa-strategically closed.

The last closure property I want to use here is {<}\kappa-distributive, which means that any \prec-chain of maximal antichains has a \prec-lower bound, where A\prec B iff every a\in A is below some b\in B. By letting player I follow their winning strategy in which player II plays elements of the antichains, we see that every \kappa-strategically closed \mathbb P is also {<}\kappa-distributive. So far so good.

We can informally describe these closure properties as enforcing the poset to be tall. The in some sense orthogonal view of enforcing the poset to be slim can be formalised using antichain properties. Here \mathbb P has the \lambda-chain condition (or \lambda-cc) if every antichain of \mathbb P has size {<}\lambda.

Now the overall question is: can a poset be both tall and slim? It turns out that there are (ZFC-provable) restrictions to this. We have to exclude the “trivial case”, which is when \mathbb P is too slim, which is to say that it isn’t really branching in any significant way. At the very minimum we should thus require that \mathbb P is atomless, which is to say that every p\in\mathbb P have incompatible extensions. Basically every forcing notion satisfies this.

If we didn’t restrict to the atomless forcings then we would have anomalies like just viewing \kappa as a (non-atomless) forcing, which trivially is both <\kappa-directed closed and also has the 2-cc. We then get our first restriction.

Proposition (folklore). Let \kappa be uncountable regular. Then every atomless \kappa-strategically closed \mathbb P does not have the \kappa-cc.

Proof. Fix a winning strategy \sigma for player I in the game. We are going to construct by simultaneous recursion two sequences \langle p_\alpha\mid\alpha<\kappa\rangle and \langle q_{\alpha+1}\mid\alpha<\kappa\rangle such that

  1. \langle p_\alpha\mid\alpha<\kappa\rangle is a decreasing sequence
  2. \langle q_{\alpha+1}\mid\alpha<\kappa\rangle is an antichain
  3. q_{\alpha+1}<p_\alpha holds for all \alpha<\kappa
  4. the p_\alpha‘s are player I’s \sigma-moves in a play of the game

For the base case simply set p_0:=1_{\mathbb P}. If p_\alpha has been defined such that \langle p_\xi\mid\xi\leq\alpha\rangle is the sequence of player I’s moves in a play of the game, then let p_{\alpha+1} be player I’s \sigma-response to whatever player II plays after p_\alpha in the game. We can use that \mathbb P is atomless to fix some q_{\alpha+1}<p_\alpha incompatible with p_{\alpha+1}.

If \langle p_\xi\mid\xi<\alpha\rangle and \langle q_{\xi+1}\mid\xi<\alpha\rangle have been defined for some limit \alpha<\kappa then use \sigma on the play thus far to get p_\alpha. At the end of the construction, \langle q_{\alpha+1}\mid\alpha<\kappa\rangle is then an antichain of size \kappa, showing that \mathbb P does not have the \kappa-cc. QED

We can get a lot of closure if we allow \mathbb P to be a bit wider:

Fact (folklore). If \kappa=\kappa^{<\kappa} then \kappa-Cohen forcing \text{Add}(\kappa,1) is both {<}\kappa-directed closed and has the \kappa^+-cc.

Proof. For the closure we simply take the union of any directed system of size {<}\kappa, which works as \kappa=\kappa^{<\kappa} implies that \kappa is regular. The forcing has size \kappa^{<\kappa}=\kappa, so it has the \kappa-cc. QED

So these previous two results give an idea of what happens to all the properties with the ‘closed’ adjective in them. But distributivity still remains, and indeed, the scenario is different here, by the following.

Fact (folklore). Forcing with a \kappa-Suslin tree is both {<}\kappa-distributive and has the \kappa-cc.

Proof. By definition of a Suslin tree, it has the \kappa-cc, and every \kappa-tree has height \kappa and is thus {<}\kappa-distributive. QED

Since we can always force a \kappa-Suslin tree on inaccessible \kappa (or \kappa=\omega_1) we get that we can consistently get a forcing satisfying both of these properties. This is as far as we can go however:

Proposition (folklore). Let \kappa be uncountable regular. Then every atomless {<}\kappa-distributive \mathbb P does not have the \lambda-cc for any \lambda<\kappa.

Proof. We recursively build a coherent sequence \langle A_\alpha\mid\alpha\leq\lambda\rangle of maximal antichains of \mathbb P. Let A_0:=\{1_{\mathbb P}\}. Assuming A_\alpha=\{a_\xi\mid\xi<\eta\} has been built (for some \eta), we use that \mathbb P is atomless to find incompatible p_\xi^{(0)},p_\xi^{(1)}\leq a_\xi. Then set A_{\alpha+1}:=\{p_\xi^{(0)},p_\xi^{(1)}\mid\xi<\eta\}. At limit stages \delta\leq\lambda we use {<}\kappa-distributivity of \mathbb P to define A_\delta. Note that |A_\alpha|\geq|\alpha| for all \alpha\leq\lambda, so A_\lambda is an antichain of size at least \lambda, showing that \mathbb P can’t have the \lambda-cc. QED

So we get that the following overview of the compatibility of these forcing properties, along with some examples of forcing posets in each category. A high-quality pdf can (as always) be found in my diagrams section.forcing_compatibility