Forcing with a Souslin tree makes $\mathfrak p=\omega_1$

I was meaning to include a proof of Farah’s lemma in my previous post, but then I realized that the slick proof assumes some background which may worth spelling out, first. Therefore, I am dedicating a short post for a self-contained proof of this lemma.

Recall that a Souslin tree is a poset $\mathbb T=\langle T,<\rangle$ satisfying the following:

1. for every $x\in T$, the set $x_\downarrow:=\{ y\in T\mid y< x\}$ is well-ordered by $<$. The set $T_\alpha:=\{ x\in T\mid \text{otp}(x_\downarrow,<)=\alpha\}$ is often considered as the $\alpha_{th}$-level of $\mathbb T$;
2. $\mathbb T$ has no uncountable antichains. In particular, $T_\alpha$ is at most countable for all $\alpha$;
3. $\mathbb T$ has no uncountable chains. In particular, $T_{\omega_1}=\emptyset$;
4. $T_0$ is a singleton;
5. $\mathbb T$ is well-pruned, that is, the set $B=\{ x\in T\mid x^{\uparrow}\text{ is countable}\}$ is empty.
   (note that clause (2) entails that $B$ is countable, so we can always pass to $T\setminus B$)

Lemma 1. Forcing with a Souslin tree $\mathbb T=\langle T,<\rangle$ adds an uncountable chain to $\mathbb T$.
Proof. Let $G$ be $\mathbb T$-generic. For all $\alpha<\omega_1$, $G$ must hit the set $D_\alpha:=\bigcup_{\alpha<\beta<\omega_1}T_\beta$. Thus, $G$ is an uncountable chain. $\square$

Lemma 2. Any Souslin tree $\langle T,<\rangle$ is isomorphic to $\langle\mathcal T,\subset\rangle$, for some $\mathcal T\subseteq{}^{<\omega_1}\omega$.
Proof. For all $\alpha<\omega_1$, fix an injection $f_\alpha:T_\alpha\rightarrow\omega$. Then, for all $x\in T$, let $\sigma_x:\text{otp}(x_\downarrow,<)\rightarrow\omega$ be the unique function that satisfies for all $\alpha\in\text{otp}(x_\downarrow,<)$: $$f_{\alpha}[T_\alpha\cap x_{\downarrow}]=\{\sigma_x(\alpha)\}.$$It is easy to see that $x<y$ iff $\sigma_x\subset\sigma_y$. Thus, letting $\mathcal T:=\{ \sigma_x\mid x\in T\}$, we get that $\langle T,<\rangle$ and $\langle\mathcal T,\subset\rangle$ are order-isomorphic. $\square$

For subsets $A,B$ of $\omega$, let $A\supset^*B$ assert that $A\setminus B$ is infinite, while $B\setminus A$ is finite.

Lemma 3. Suppose that $\mathcal T\subseteq{}^{<\omega_1}\omega$ is a Souslin tree (under the $\subset$ relation).
Then there exists an injection $\psi:\mathcal T\rightarrow[\omega]^\omega$ such that:

• $\psi$ witnesses that $\langle\mathcal T,\subset\rangle$ is order-isomorphic $\langle\psi[\mathcal T],\supset^*\rangle$;
• for every $\alpha<\omega_1$, and $\sigma,\eta\in T_\alpha$, we have $\psi(\sigma)\cap\psi(\eta)=\emptyset$.

Proof. For every limit nonzero ordinal $\alpha$, let $\{ \alpha_n\mid n<\omega\}$ be the increasing enumeration of some fixed cofinal subset of $\alpha$, and let $f_\alpha:|T_\alpha|\rightarrow T_\alpha$ be some bijection.
We now define $\psi:\mathcal T\rightarrow[\omega]^\omega$ by recursion on the levels.

• Let $\psi(\emptyset):=\omega$;
• Suppose that we are given $\sigma\in T_\alpha$, where $\psi\restriction T_\alpha$ has already been defined. Then, fix a partition $\{ A_n\mid n<\omega\}$ of $\psi(\sigma)$ into infinite sets, and let $\psi(\sigma{}^\frown n):=A_n$ for all $n<\omega$ such that $\sigma{}^\frown n\in T_{\alpha+1}$;
• Suppose that $\alpha<\omega_1$ is nonzero limit ordinal, and $\psi\restriction T_\beta$ has been defined for all $\beta<\alpha$. We shall define $\psi(\sigma)$ for all $\sigma\in T_\alpha$ by recursion on the well-ordering of $T_\alpha$ as dictated by $f_\alpha$. For $\sigma=f_\alpha(0)$, we simply find (e.g., via a recursive construction) an injection $i_0:\omega\rightarrow\omega$ such that $i_0(n)\in\bigcap_{k\le n}\psi(\sigma\restriction\alpha_k)$ for all $n<\omega$, and then put $\psi(\sigma):=i_0[\omega]$. Then, for $\sigma=f_\alpha(m+1)$, we pick an injection $i_{m+1}:\omega\rightarrow\omega$  such that $$i_{m+1}(n)\in\bigcap_{k\le n}\psi(\sigma\restriction\alpha_k)\setminus\bigcup\{\psi(f_\alpha(j))\mid j\le m\}$$ for all $n<\omega$, and then, put $\psi(\sigma):=i_{m+1}[\omega]$. $\square$

Corollary A. Any Souslin tree is isomorphic to $\langle\mathcal T,\supset^*\rangle$, for some $\mathcal T\subseteq[\omega]^\omega$, in which for every $\alpha<\omega_1$, the level $T_\alpha$ consists of mutually disjoint sets.

Lemma 4. Any Souslin tree $\mathbb T=\langle T,<\rangle$ is an $\omega$-distributive notion of forcing, that is, if $\langle D_n\mid n<\omega\rangle$ is a sequence of cofinal upward-closed subsets of $\mathbb T$, then $\bigcap_{n<\omega}D_n\neq\emptyset$.
Proof. For every $n<\omega$, let $$A_n:=\{ x\in D_n\mid x_\downarrow\cap D_n=\emptyset\}.$$ Then $A_n$ is an antichain, and hence countable. Let $$\alpha:=\sup\left\{ \text{otp}(x_\downarrow,<)\mid x\in\bigcup_{n<\omega}A_n\right\}.$$Then $\alpha<\omega_1$. Pick $x\in T_{\alpha+1}$. We claim that $x\in\bigcap_{n<\omega}D_n$.
To see this, fix an arbitrary $n<\omega$. As $D_n$ is cofinal, let us pick some $y_n\in D_n$ such that $x<y_n$. As $A_n$ is the set of minimal elements of $D_n$, we may also find $x_n\in A_n$ such that $x_n<y_n$. So $\{ x_n,x\}\subseteq (y_n)_\downarrow$ and hence either $x_n<x$ or $x<x_n$. Recalling that $x\in T_{\alpha+1}$ and the definition of $\alpha$, we conclude that $x_n<x<y_n$. Recalling that $x_n\in D_n$ and $D_n$ is downward closed, we have $x\in D_n$. $\square$

Corolloary B. Forcing with a Souslin tree $\mathbb T$ adds no infinite subsets to $\omega$.
Proof. Suppose that $\sigma$ is a $\mathbb T$-name for a function from $\omega$ to $\omega$. For all $n<\omega$, the set $D_n$ of all $\mathbb T$-conditions that decides $\sigma(n)$ is cofinal and upward-closed. By Lemma 3, then, we find a condition that decides $\sigma(n)$ for all $n<\omega$, and hence forces that $\sigma$ coincides with a ground model function from $\omega$ to $\omega$. $\square$

Theorem (Farah, 1996). If $\mathbb T$ is a Souslin tree, then $V^{\mathbb T}\models\mathfrak p=\omega_1$.
Proof. By Corollary A, we may assume that $\mathbb T=\langle \mathcal T,\supset^*\rangle$, where $\mathcal T\subseteq[\omega]^\omega$, and each of the levels of $\mathbb T$ consists of mutually disjoint sets. By Lemma 1, let $\mathcal C\subseteq\mathcal T$ be, in the extension, some uncountable $\supset^*$-chain. By passing to $\mathcal C_\downarrow$, we may assume that $\mathcal C$ is maximal, that is, $\mathcal C\cap T_\alpha\neq\emptyset$ for all $\alpha<\omega_1$.
To see that $\mathfrak p=\omega_1$, it now suffices to argue that $\mathcal C$ admits no infinite pseudointersection.
Towards a contradiction, suppose that $P\in[\omega]^\omega$ is a pseudointersection for $\mathcal C$. Put $\mathcal C’:=\{ A\in\mathcal T\mid A\supset^* P\}$. By Corollary B, the collection $\mathcal C’$ lies in the ground model, and so to reach a contradiction, we argue that $\mathcal C’=\mathcal C$.
It is clear that $\mathcal C\subseteq\mathcal C’$. Next, suppose that $A\in \mathcal C’$. Let $\alpha<\omega_1$ be such that $A\in T_\alpha$.  Pick and $B\in\mathcal C\cap T_\alpha$. Then $A\supset^* P, B\supset^* P$, and hence $A\cap B\neq\emptyset$, which means that $A=B$ and hence $A\in\mathcal C$. $\square$