No new real is contained in a countable closed set

Question

I am trying to read Michael Hrusák's "MAD Families and the rationals" and I have studied Forcing using Kunen's books. On theorem 1, the author says that no new real is contained in a countable closed set coded in the ground model. If I understood it right, what the author stated is that if $M$ is a ctm and $A\subset (2^\omega)^M$ is such that $A \in M$ and $\operatorname{cl}^MA$ is countable , then $\operatorname{cl}^{M[G]}A=\operatorname{cl}^{M}A$. My question is: Why?

Answer 1

While you can code a closed set $A \subseteq 2^\omega$ by any countable dense subset of $A$; I think it is more transparent to code $A$ by the tree $T_A:=\{ f\upharpoonright n : f\in A, n\in \omega\}$. Then $A=[T_A] = \{f: \forall n\ f\upharpoonright n \in T\}$.

Any countable closed set $A$ can be decomposed by Cantor-Bendixson derivatives as follows:

• Let $T_0 = T_{A}$, $A_0 = A$, and let $B_0$ be the set of isolated points of $A_0$ (which can be found as the isolated branches of $T_0$).
• Set $T_1 = $ the nodes of $T_0$ which have at least two extensions in $A_0$, and $A_1 = A_0 \setminus B_0$.
• Then $A_1 = [T_1]$ = the set of non-isolated points of $A$ = the "first CB derivative" of $A$.
• Define $A_2$, $T_2$ similarly. etc.
• Let $T_\omega$ be the intersection of the decreasing sequence $T_n$, and continue by letting $T_{\omega+1}$ be the derivative of $T_\omega$.
• Continue by transfinite induction. Eventually (after countably many steps) you will reach the empty set.
• You have now written $A$ as the disjoint union of set $B_0 \cup B_1\cup \cdots$.

The construction of the sequence $T_\alpha$ is absolute (as the definition off the derivative is arithmetical, it used only quantifications over the countable set $2^{<\omega}$). So it will give the same results in any model.

If you compare $V$ and $V[G]$, the sets $B_0$ defined in these models (isolated branches of $T_0$) will be the same. But then also the sets $B_1$ etc will be the same in both models. etc.

Answer 2

Let $P$ be a closed set with code in $V$ and $P$ is countable in $V[G]$ where $G$ is generic for some forcing $\mathbb{P}$. This means that there is some tree $T$ consisting of binary sequences so that $P = [T]$, where the latter is the set of infinite paths through $T$, and $T$ belongs to the ground model $V$.

First the claim is that $[T]$ is countable in $V$. If not, by the perfect set property for closed set (which you can prove by the Cantor-Bendixson idea of the other answer), $[T]$ has a perfect subset. This means there is a subtree $S \subseteq T$ with the property that for all nodes $s \in S$, there is a extension $t \supseteq s$ so that $t0, t1 \in S$. ($S$ is a perfect tree.)

Go back into $V[G]$, $S \subseteq T$ so $[S] \subseteq [T] = P$. But since $S$ is a perfect tree, $[S]$ must be in bijection with $2^\omega$. So $P$ is not countable. Contradiction.

This shows that $[T]$ is countable in $V$. So pick some $f : \omega \rightarrow [T]$ which is a bijection of $\omega$ onto $[T]$ and $f$ belongs to $V$. $V$ satisfies that statement $$(\forall x \in 2^{\omega})(x \in [T] \Rightarrow (\exists n \in \omega)(f(n) = x))$$ This is a $\Pi_1^1$ statement of descriptive set theory. By Mostowski absoluteness, this statement remains true in $V[G]$. Hence in $V[G]$, one can show $[T] \subseteq f''\omega = [T]^V$. The latter consists of only elements of $V$.

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Another way to see this:

If $A$ is closed, Borel, or analytic in code $x$ and $A$ is countable, then $A$ consists entirely of $x$-hyperarithmetic elements. This means, let $\omega_1^x$ be the sup of $x$-computable ordinals. Then $A \subseteq L_{\omega_1^x}[x]$, a level of Godel constructible hierarchy.

In $V[G]$, if $A$ is analytic in parameter $x \in V$ and $A$ is countable, then $A \subseteq L[x] \subseteq V$. So $A$ has only ground model elements.