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 $\mathbb{T}=⟨T,<⟩$ 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, $T_{{\omega }_1}=\mathrm{\varnothing }$;
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 $\mathbb{T}=⟨T,<⟩$ 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. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Lemma 2. Any Souslin tree $⟨T,<⟩$ is isomorphic to $⟨\mathcal{T},\subset ⟩$, for some $\mathcal{T}\subseteq {}^{<{\omega }_1}\omega$.
Proof. For all $\alpha <{\omega }_1$, fix an injection $f_{\alpha }:T_{\alpha }\to \omega$. Then, for all $x\in T$, let ${\sigma }_x:\text{otp}(x_{\downarrow },<)\to \omega$ 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 $⟨T,<⟩$ and $⟨\mathcal{T},\subset ⟩$ are order-isomorphic. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

For subsets $A,B$ of $\omega$, let $A{\supset }^{\ast }B$ 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 $\psi :\mathcal{T}\to [\omega {]}^{\omega }$ such that:

• $\psi$ witnesses that $⟨\mathcal{T},\subset ⟩$ is order-isomorphic $⟨\psi [\mathcal{T}],{\supset }^{\ast }⟩$;
• for every $\alpha <{\omega }_1$, and $\sigma ,\eta \in T_{\alpha }$, we have $\psi (\sigma )\cap \psi (\eta )=\mathrm{\varnothing }$.

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 $f_{\alpha }:|T_{\alpha }|\to T_{\alpha }$ be some bijection.
We now define $\psi :\mathcal{T}\to [\omega {]}^{\omega }$ by recursion on the levels.

• Let $\psi (\mathrm{\varnothing }):=\omega$;
• Suppose that we are given $\sigma \in T_{\alpha }$, where $\psi \upharpoonright T_{\alpha }$ has already been defined. Then, fix a partition $\{A_n\mid n<\omega \}$ of $\psi (\sigma )$ into infinite sets, and let $\psi (\sigma {}^{⌢}n):=A_n$ for all $n<\omega$ such that $\sigma {}^{⌢}n\in T_{\alpha +1}$;
• Suppose that $\alpha <{\omega }_1$ is nonzero limit ordinal, and $\psi \upharpoonright T_{\beta }$ 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 $i_0:\omega \to \omega$ such that $i_0(n)\in \bigcap _{k\le n}\psi (\sigma \upharpoonright {\alpha }_k)$ for all $n<\omega$, and then put $\psi (\sigma ):=i_0[\omega ]$. Then, for $\sigma =f_{\alpha }(m+1)$, we pick an injection $i_{m+1}:\omega \to \omega$  such that $$i_{m+1}(n)\in \bigcap _{k\le n}\psi (\sigma \upharpoonright {\alpha }_k)\setminus \bigcup \{\psi (f_{\alpha }(j))\mid j\le m\}$$ for all $n<\omega$, and then, put $\psi (\sigma ):=i_{m+1}[\omega ]$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary A. Any Souslin tree is isomorphic to $⟨\mathcal{T},{\supset }^{\ast }⟩$, 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 $\mathbb{T}=⟨T,<⟩$ is an $\omega$-distributive notion of forcing, that is, if $⟨D_n\mid n<\omega ⟩$ is a sequence of cofinal upward-closed subsets of $\mathbb{T}$, then $\bigcap _{n<\omega }{D}_n\ne \mathrm{\varnothing }$.
Proof. For every $n<\omega$, let $$A_n:=\{x\in D_n\mid x_{\downarrow }\cap D_n=\mathrm{\varnothing }\}.$$ Then $A_n$ is an antichain, and hence countable. Let $$\alpha :=sup\{\text{otp}(x_{\downarrow },<)\mid x\in \bigcup _{n<\omega }{A}_n\}.$$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 $x\in D_n$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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 $\omega$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

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 $\mathbb{T}=⟨\mathcal{T},{\supset }^{\ast }⟩$, 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 ${\supset }^{\ast }$-chain. By passing to ${\mathcal{C}}_{\downarrow }$, we may assume that $\mathcal{C}$ is maximal, that is, $\mathcal{C}\cap T_{\alpha }\ne \mathrm{\varnothing }$ 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 ${\mathcal{C}}^{\prime }:=\{A\in \mathcal{T}\mid A{\supset }^{\ast }P\}$. By Corollary B, the collection ${\mathcal{C}}^{\prime }$ lies in the ground model, and so to reach a contradiction, we argue that ${\mathcal{C}}^{\prime }=\mathcal{C}$.
It is clear that $\mathcal{C}\subseteq {\mathcal{C}}^{\prime }$. Next, suppose that $A\in {\mathcal{C}}^{\prime }$. Let $\alpha <{\omega }_1$ be such that $A\in T_{\alpha }$.  Pick and $B\in \mathcal{C}\cap T_{\alpha }$. Then $A{\supset }^{\ast }P,B{\supset }^{\ast }P$, and hence $A\cap B\ne \mathrm{\varnothing }$, which means that $A=B$ and hence $A\in \mathcal{C}$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$