Suslin trees in transitive models

Question

Let $M$ be a transitive model of $ZFC$ with $\omega_1\in M$ such that $M\models ((T,<)\text{ is a Suslin tree})$. Fix $A\in M$ such that $M\models (A\text{ is a maximal antichain})$ and $b\subset T$ an $\omega_1$-branch. I'd like to show that $b\cap A\neq \emptyset$.

Some things that seem true:

• Since $\omega_1\in M$, things are countable in $V$ iff they are countable in $M$.
• The function $\alpha\mapsto \text{level}_\alpha(T)$ is absolute, i.e. $\text{level}_\alpha(T)^M=\text{level}_\alpha(T)$. In particular, $T$ has height $\omega_1$ (in $V$).
• Being a chain/antichain is absolute.
• By the above, we know that all chains, antichains and levels are countable. In particular, $A$ is countable and thus contained in some countable union of levels.
• If $b=\{x_\xi:\xi<\omega_1\}$, using the fact that $A$ is a maximal antichain, for each $\xi<\omega_1$ there is some $y_\xi\in A$ such that $x_\xi$ and $y_\xi$ are comparable. By the pigeonhole principle, the same $y$ has to work for $\aleph_1$ many $\xi$, and by a further application of this and the fact that the tree has height $\omega_1$, there is some $y\in A$ below $\aleph_1$ many elements of $b$.

Could I get a hint on where to go from here? I realize that my latter remarks do not mention $M$, so I'm thinking the model should somehow come into the discussion, but I don't see how.

Answer 1

You're basically moving in the right direction. A couple comments:

• If you can show that some element of $A$ is extended by any element of $b$, you're done: $b$ is a branch, which means it's closed downwards, so this would give $b\cap A\not=\emptyset$.

• So now you just have to make sure you've really shown this. As you observe, you haven't mentioned $M$, so something's being swept under the rug here. The key point is: can you argue that $A$ is actually a maximal antichain in $T$? $M$ certainly thinks it is, but it could be wrong about this in principle. (HINT: use the transitivity of $M$ ...)

• And at this point your final paragraph is actually doing too much work. Instead, just note that since $A$ is countable and maximal, there is some level $\alpha<\omega_1$ such that every element of $A$ has height $<\alpha$; now, do you see why the $\alpha$th element of $b$ must extend some element of $A$ in order for $A$ to be (as you will have shown) a genuinely maximal antichain?