cardinality of perfect sets in generalized Baire space

Question

I've been unable to find an answer to the following question in the literature on generalized descriptive set theory. Consider Baire space $\kappa^{\kappa}$ where $\kappa$ is inaccessible. The basic open sets are the $U_f$ where $f\in\kappa^{<\kappa}$. A perfect set is a nonempty closed set with no isolated points. Does a perfect set have cardinality $2^{\kappa}$? Does it have to have cardinality $>\kappa$?

A variation of an observation in "Generalized Descriptive Set Theory and Classification Theory" (arxiv.org/abs/1207.4311) states that if $T$ is a slim Kurepa tree (as defined in Devlin's "Constructibility") then there is no continuous (in fact Borel) injection of $2^{\kappa}$ into $[T]$, but this doesn't answer my question.

Of course, if the existence of a slim Kurepa tree was consistent with $2^{\kappa}>\kappa^+$ then it would be consistent that my first question had a "no" answer; I have not been able to find any references on this question, but I wouldn't be surprised if it were so.

Answer 1

It is consistent to have a perfect set of size $\kappa$, or of intermediate size between $\kappa$ and $2^\kappa$.

To see this, suppose that $\kappa$ is an inaccessible cardinal in $V$, and let $T=2^{<\kappa}$ be the tree of ${<}\kappa$ binary sequences, and let $X=[T]={}^\kappa 2$ be the set of branches through this tree in $V$. So $X$ is a perfect set in $V$ and has size $2^\kappa$ there.

Next, force over $V$ to add a Cohen real $c\subset\omega$ and then to collapse $(2^\kappa)^V$ to $\kappa$. Consider the resulting forcing extension $V[c][G]$.

The combined forcing admits a closure point at $\omega_1$, and it follows from lemma 13 of my article Extensions with the approximation and cover properties have no new large cardinals that the extension $V[c][G]$ does not add any new branches through a tree of height $\kappa$ in the ground model $V$. In particular, $X$ is still the set of all branches through $T$ in $V[c][G]$. Note that $X$ still has no isolated points. Further, since there are no new branches through $T$ in $V[c][G]$, it follows that $X$ is still closed (since any new element of the closure of $X$ would have to be a branch through $T$).

So $X$ is perfect in $V[c][G]$, but since $(2^\kappa)^V$ was collapsed to $\kappa$, it follows that $X$ has size $\kappa$ there.

One can modify the argument to make a perfect set of size between $\kappa$ and $2^\kappa$. Start in $V$ with $2^\kappa=\kappa^+$, but now force to add a Cohen real and then let $G$ pump up $2^\kappa$ very high. The set $X=({}^\kappa 2)^V$ will still be perfect in $V[c][G]$, but now it will have size $\kappa^+$, even though $2^\kappa$ is large in $V[c][G]$. One can make it have size $\kappa^{++}$ or whatever you like, in the same way.

Answer 2

For every $\kappa$ of uncountable cofinality there is a tree $T\subseteq 2^{<\kappa}$ such that $[T]=\kappa$. The tree $T$ is the tree of all binary sequences $f\colon \alpha \to 2$, $\alpha <\kappa$ such that $f^{-1} (1)$ is finite. It is clear that for every $f \in T$, $f\frown (1), f\frown (0)$ are both in $T$, so this tree is prefect.

Any branch $b$ of $T$ contains finitely many $1$-s, since $\text{cf }\kappa > \omega$ and therefore is there were infinitely many $1$-s, there was some $\alpha$ such that $b\restriction \alpha$ contains already infinitely many $1$-s. On the other hand, it is clear that for every $b\colon \kappa \to 2$, with $b^{-1}(1)$ finite, $\{ b\restriction \alpha \mid \alpha < \kappa \}$ is a branch in $T$.

There are $\kappa^{<\omega} = \kappa$ such branches, as wanted.

Edit: I argue that the $ZFC$ doesn't prove that there are prefect sets in $^\kappa\kappa$ of size $\kappa^{+}$. The proof is similar to the proof of the consistency of "there are no Kurepa trees".

Theorem: Assume $GCH$. Let $\kappa < \eta < \mu$ be regular cardinals, $\eta$ inaccessible. Then after forcing with $\mathbb{Q} = Add(\kappa,\mu)\times Col(\kappa,<\eta)$, for every $\kappa$-tree $T$, $|[T]| \in \kappa \cup \{\kappa, \mu\}$. In this generic extension $\eta = \kappa^+$, $\mu = 2^\kappa$ and every cardinal $\geq \eta$ is preserved.

Proof: We need the following well known fact:

Fact: Let $T$ be a tree of height $\kappa$. If there is a $\kappa$-closed forcing that adds a branch to $T$ then $|[T]| = 2^\kappa$.

Sketch of proof: Let $\mathbb{P}$ be a $\kappa$-closed forcing that adds a branch for $T$ and let $\dot{b}$ be the name of this new branch. Define an embedding of $2^{<\kappa}$ into $T$ by building a tree of conditions in $\mathbb{P}$, $\langle p_\eta \mid \eta \in 2^{<\kappa}\rangle$ such that for every $\eta$, $p_{\eta \frown (0)}, p_{\eta \frown (1)} \leq p$ give contradictionary information about the branch. Then for every $f\in 2^\kappa$, $b_f = \{ t\in T \mid \exists \alpha < \kappa,\,p_{f\restriction \alpha}\Vdash t\in \dot{b}\}$ is a cofinal branch, and $f\neq g\implies b_f \neq b_g$. Q.E.D.

Let's return to the proof of the theorem. Let $G$ be a $\mathbb{Q}$-generic filter and let $\dot{T}$ be a $\mathbb{Q}$-name for a tree in $V[G]$. Note that $(\kappa^{<\kappa})^{V[G]} = (\kappa^{<\kappa})^{V}$, so we may assume that $\Vdash \dot{T}\subseteq \check{\kappa^{<\kappa}}$.

By the $\eta$.c.c. of $\mathbb{Q}$, we can find a model $M\prec H_\chi$ such that $|M|<\eta$, $\dot{T}, \mathbb{Q} \in M$, $^{<\kappa}M\subseteq M$ and for every $t\in \kappa^{<\kappa}$ there is a maximal antichain $\mathcal{A} \subseteq M$ that decides whether $t\in \dot{T}$ or not (so $T\in V[M\cap G]$).

Note that $M\cap G$ is the restriction of the generic filter to the coordinates that are ordinals of $M$, so it is a generic filter for the forcing $\mathbb{Q}_M := Add(\kappa, \mu\cap M)\times Col(\kappa, <(\eta\cap M))$. Let $\mathbb{P}$ be the restriction of $Add(\kappa,\mu)\times Col(\kappa,<\eta)$ to the ordinals that don't appear in $M$, so $\mathbb{Q}= \mathbb{Q}_M \times \mathbb{P}$.

In $V[G\cap M]$, $2^\kappa$ is less than $\eta$ (since $|\mu \cap M| < \eta$), so if all the branches of $T$ in $V[G]$ are already in $V[G\cap M]$, we have that $V[G]\models |[T]|< \eta = \kappa^+$, and we're done.

Otherwise, let $\dot{b}$ be a name for a new branch. Since $\mathbb{P} \cong \mathbb{P} \times \mathbb{P}$, $\mathbb{Q} \cong \mathbb{Q}_M \times \mathbb{P} \times \mathbb{P}$. Therefore also in $V[G]$, $\dot{b}$ is a $\mathbb{P}$-name for a new branch (by the mutually generity of the two copies of $\mathbb{P}$). Moreover, $\mathbb{P}$ is $\kappa$-closed in $V[G]$, so by the fact above - $V[G] \models |[T]|=2^\kappa$, as wanted.