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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.

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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.

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  • $\begingroup$ It is not clear to me whether one might be able to prove that there is always a perfect set of size $\kappa$. $\endgroup$ Sep 27 '14 at 21:25
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    $\begingroup$ I think that the set of branches of the tree $T\subseteq 2^{<\kappa}$, that contains only sequences with finitely many ones, is a perfect set of size $\kappa$ (since $\text{cf }\kappa > \omega$, there are $\kappa^{<\omega}$ branches). $\endgroup$
    – Yair Hayut
    Sep 28 '14 at 7:18
  • $\begingroup$ @Yair I think that is right! Please post an answer. $\endgroup$ Sep 28 '14 at 10:58
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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.

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  • $\begingroup$ A minor quibble: unlike the situation in Cantor space, a perfect set in $2^\kappa$ is not the same thing as the set of branches through a splitting tree on $2^{<\kappa}$, since what you need to know is not just that the tree splits, but also that those splitting nodes lie on branches. (For example, imagine an almost-Aronszajn tree, which is splitting, but has just one or just a few branches). But it is not problem for you, since every node lies on the branch that continues with all zeros. $\endgroup$ Sep 28 '14 at 17:00
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    $\begingroup$ @Joel: You're right. I wonder, is it possible to show (in a similar way?) that every cardinal between $\kappa$ and $2^\kappa$ is the cardinality of some perfect set in $2^\kappa$? $\endgroup$
    – Yair Hayut
    Sep 28 '14 at 18:23
  • $\begingroup$ I wonder that also. My answer shows at least that this is consistent with $2^\kappa$ being large. $\endgroup$ Sep 28 '14 at 18:34
  • $\begingroup$ I just found that this construction appears "A Cantor Bendixson theorem for the space ${}^{\omega_1}\omega_1$" by J. Vaananen. $\endgroup$ Nov 17 '14 at 4:37
  • $\begingroup$ @MartinDowd: This sounds reasonable. Also Jin and Shelah have similar and even stronger theorems, see arxiv.org/abs/math/9211214. $\endgroup$
    – Yair Hayut
    Nov 17 '14 at 6:05

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