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The generalized Cantor space is the space $2^\kappa$, with basic open sets $$ [\sigma] := \{f\in 2^\kappa : \sigma\subseteq f\}, $$ for $\sigma\in 2^{<\kappa}$.

A space is $\kappa$-compact if every open cover has a subcover of cardinality (strictly) smaller than $\kappa$.

Problem. For which cardinals $\kappa$ does the generalized Cantor space $2^\kappa$ embed as a subspace of every $\kappa$-compact set $C\subseteq 2^\kappa$ with $|C|>\kappa$?

It is a classic result that this holds for $\kappa=\omega$.

In the original formulation of this problem, I mentioned that, in light of an earlier answer, the case where $\kappa$ is weakly compact is particularly interesting, and the expected answer is "for all of these". According to Yair Hayut's answer below, this is wrong. The full problem, as stated above, remains open (we need cardinals for which the answer is positive).

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  1. Let $\kappa$ be an uncountable regular cardinal that is not weakly compact and let $C$ be a $\kappa$-compact subset of ${}^\kappa 2$. Assume, towards a contradiction, that there is a continuous injection of ${}^\kappa 2$ into $C$. Since $C$ is closed in ${}^\kappa\kappa$, Lemma 2.9 of this paper shows that $C$ contains a closed set homeomorphic to ${}^\kappa 2$. By our assumption, $\kappa$ is not weakly compact and a theorem of Hung and Negrepontis (see Corollary 2.3 of the above paper) implies that the spaces ${}^\kappa 2$ and ${}^\kappa\kappa$ are homeomorphic. This is a contradiction, because an easy argument shows that $\kappa$-compact subsets of ${}^\kappa\kappa$ do not contain closed subsets homeomorphic to ${}^\kappa\kappa$ (see Fact 1.5 of the above paper).

  2. Let $\kappa$ be indestructible supercompact, let $\lambda>\kappa$ be inaccessible and let $G$ be $Col(\kappa,{<}\lambda)$-generic over $V$. In this situation, standard arguments (see, for example, Lemma 7.6 of this paper) show that, in $V[G]$, all subtrees of ${}^{{<}\kappa}\kappa$ with more than $\kappa$-many $\kappa$-branches contain a subtree isomorphic to ${}^{{<}\kappa}2$. In particular, in $V[G]$, $\kappa$ is weakly compact and every $\kappa$-compact subset of ${}^\kappa 2$ of cardinality greater than $\kappa$ contains a continuous injective image of ${}^\kappa 2$.

  3. Let $\kappa$ be a weakly compact cardinal with the property that there is a subset $A\subseteq\kappa$ such that $\kappa^+=(\kappa^+)^{L[A]}$ and the set $\{\alpha\in S^\kappa_\omega \mid cof^{L[A\cap\alpha]}(\alpha)=\omega\}$ is stationary in $\kappa$. In this situation, a modification of a classical argument of Solovay shows that there is a weak $\kappa$-Kurepa tree, i.e. a subtree $T$ of ${}^{{<}\kappa}2$ with $\kappa^+$-many $\kappa$-branches and $\vert T(\alpha)\vert=\vert\alpha\vert$ for stationary many $\alpha<\kappa$. Since $\kappa$ is weakly compact, the tree $T$ contains no $\kappa$-Aronszajn subtrees and a theorem of Juhász and Weiss (see Lemma 2.2 of the first paper) shows that the corresponding closed subset $[T]$ of ${}^\kappa 2$ is $\kappa$-compact. By results of Mekler and Väänänen, there is no continuous embedding of ${}^\kappa 2$ into $[T]$.

  4. The existence of weak $\kappa$-Kurepa trees at every inaccessible cardinal $\kappa$ is consistent with the existence of very large large cardinals (including supercompact cardinals). This is discussed on page 33 of this paper by S. Friedman, Hyttinen and Kulikov.

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  • 2
    $\begingroup$ If $T$ is a tree that contains a subtree isomorphic to $2^{<\kappa}$ then $\text{Add}(\kappa, 1)$ adds to $[T]$ a new element. Thus, if you just want to get the consistency of having a many large cardinals but for every weakly compact, $\kappa$, a $\kappa$-tree $T$ such that $2^{<\kappa}$ does not embed into it - you can start with a model with many large cardinals and add a single Cohen real. In the generic extension, $2^{<\kappa}$ is not a subtree of any ground model $\kappa$-tree, by Hamkin's Gap argument. $\endgroup$ – Yair Hayut Apr 17 '16 at 15:31
  • $\begingroup$ @Philipp: I would appreciate some additional clarifications: 1. Remark (1) on page 33 of the paper you cited in your item 4: In $L$, is there a weak Kurepa tree for all uncountable regular (inaccessible?) $\kappa$? 2. Does your item 3 assert that the existence of a weak Kurepa tree for $\kappa$ imply $2^\kappa$ does not embed in every $\kappa$-compact space? $\endgroup$ – Boaz Tsaban May 16 '16 at 13:17
  • $\begingroup$ 1. The existence of weak Kurepa trees at all uncountable regular in L already follows from Solovay's argument mentioned in item 3. The cited paper sketches a class forcing construction that adds a weak Kurepa tree at every inaccessible cardinal, while preserving many large cardinals. 2. Yes. This short argument is, for example, contained in Section 7 of the paper cited in item 2. $\endgroup$ – Philipp Lücke May 20 '16 at 17:49
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EDIT:

As Boaz has pointed out in the comments, there is a mistake in my alleged proof. After thinking about it a bit, I believe it is a genuine mistake (i.e., not quickly fixable). This error does not come in until the end of my argument (Lemma 5), so I will leave the original post up and intact in case the first part of the argument (Lemmas 1-4) can still be useful.

ORIGINAL POST:

Every weakly compact cardinal has this property.

I don't know a reference for this, but I can give a proof. The proof is just a generalization of one proof for this fact when $\kappa = \omega$.

Proposition: If $\kappa$ is weakly compact, then the generalized Cantor space $2^\kappa$ is homeomorphic to a subspace of every $\kappa$-compact subset $C$ of $2^\kappa$ with $|C| > \kappa$.

Proof: I'll try to break it up into bite-sized lemmas:

Lemma 1: Suppose $X$ is a Hausdorff space in which an intersection of fewer than $\kappa$ open sets is open. Then every $\kappa$-compact subspace of $X$ is closed. Furthermore, if $X$ is $\kappa$-compact then a subspace $C$ of $X$ is $\kappa$-compact if and only if it is closed.

Proof: (In many ways, this is just like the proof for $\kappa = \omega$.) Suppose $A \subseteq X$ is $\kappa$-compact, and fix $x \notin A$. For each $y \in A$, let $U_y$ and $V_y$ be open sets with $x \in U_y$, $y \in V_y$, and $U_y \cap V_y = \emptyset$. The collection $\{V_y : y \in A\}$ covers $A$, so by $\kappa$-compactness some subset $\mathcal{V}$ with $|\mathcal{V}|< \kappa$ still covers $A$. By our assumptions about $X$, $$\{U_y : V_y \in \mathcal{V}\}$$ is a neighborhood of $x$, and it must be disjoint from $A$. Thus $A$ is closed.

It remains to show that if $X$ is $\kappa$-compact, then every closed $C \subseteq X$ is also. If $\mathcal U$ is an open cover of $C$, then $\mathcal U \cup \{X - C\}$ is an open cover of $X$. Since $X$ is $\kappa$-compact, we may pass to a $<\kappa$-sized subcover. Removing $X-C$ from this subcover if necessary, we obtain a $<\kappa$-sized subset of $\mathcal U$ that covers $C$. $\ $QED

Lemma 2: Assume $\kappa$ is regular. In the generalized Cantor space $2^\kappa$, any intersection of fewer than $\kappa$ open sets is open.

Proof: Using regularity, it is easy to see that this is true for basic open sets. The result follows easily. $\ $QED

Putting these two lemmas together with Joseph van Name's answer to your previous question, we obtain:

Lemma 3: If $\kappa$ is weakly compact, then $C \subseteq 2^\kappa$ is $\kappa$-compact if and only if it is closed.

Lemma 4: If $\kappa$ is regular and $2^{<\kappa} = \kappa$, then every closed $C \subseteq 2^\kappa$ with $|C| > \kappa$ contains a closed subspace with no isolated points.

Proof: Fix a closed subset $C$ of $2^\kappa$.

By transfinite recursion, define a sequence of subsets of $2^\kappa$ as follows: $$C_0 = C$$ $$C_{\alpha+1} = C_\alpha'$$ $$C_{\lambda} = \bigcap_{\alpha < \lambda}C_\alpha \qquad \text{ for limit }\lambda$$ (here $C_\alpha'$ denotes the derived set, or Cantor-Bendixson derivative, of $C_\alpha$). Eventually this sequence stabilizes, and we obtain some $C_\rho \subseteq C$ with $C_\rho = C_\rho'$. Using the fact that $2^\kappa$ has a basis of size $\kappa$ (because $2^{<\kappa} = \kappa$), it is not difficult to show that $\rho < \kappa^+$ and that $|C_{\alpha+1} - C_\alpha| \leq \kappa$ for all $\alpha < \rho$. Because $|C| > \kappa$, it follows that $C_\rho \neq \emptyset$. Furthermore, since $C_\rho = C_\rho'$, $C_\rho$ has no isolated points. $\ $QED

Lemma 5: Suppose $\kappa$ is regular. Then $2^\kappa$ is the only non-empty, $\kappa$-compact, zero-dimensional Hausdorff space with a basis of size $\kappa$, no isolated points, and in which every intersection of fewer than $\kappa$ open sets is open.

Proof sketch: Let $X$ be any space fitting the above description. Let $\{B_\alpha : \alpha < \kappa\}$ be a basis for $X$ consisting of clopen sets.

We now define a map $\varphi$ from the tree $2^{<\kappa}$ into the set of clopen subsets of $X$. Begin by putting $\varphi(\emptyset) = X$. If $\alpha$ is a limit ordinal and $\varphi$ is already defined on $2^{<\alpha}$, then extend the map to $2^\alpha$ by taking intersections: $$\varphi(f) = \bigcap_{\beta < \alpha}\varphi(f|_\beta) \qquad \text{ for }f \in 2^\alpha.$$ If our map is already defined on $2^\alpha$, then define it on $2^{\alpha+1}$ by splitting each $\varphi(f)$, $f \in 2^\alpha$, into two clopen pieces. Do this in such a way that each piece is nonempty (recall there are no isolated points in $X$). Also, if $\emptyset \neq B_\alpha \cap \varphi(f) \neq \varphi(f)$, then split $\varphi(f)$ into $B_\alpha \cap \varphi(f)$ and its complement.

This tree defines a homeomorphism from $2^\kappa$ to $X$. If $f \in 2^\kappa$, our homeomorphism takes $f$ to the unique element of $\bigcap_{\alpha < \kappa}\varphi(f|_\alpha)$. This intersection is nonempty by $\kappa$-compactness, and it is a singleton by our use of the $B_\alpha$ in the successor stages of our construction. Thus the map is well-defined, and it's not too hard to check that it is a homeomorphism. $\ $QED

I admit I left out some details from this last proof. Please feel free to ask for more details if you want them.

Putting everything together, this finishes the proof of the proposition. $\ $QED

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    $\begingroup$ It is unclear how to make sure that the sets $\bigcap_{\beta < \alpha}\varphi(f|_\beta)$ are nonempty in limit stages. Hopefully some remedy can be found. $\endgroup$ – Boaz Tsaban Apr 12 '16 at 16:49
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It is possible that $\kappa$ is weakly compact (and even more) but there is a $\kappa$-compact space $C\subseteq 2^\kappa$ such that $|C| = \kappa^{+} < 2^\kappa$.

First, let us note that if $\kappa$ is weakly compact then for every tree $T\subseteq 2^{<\kappa}$, the set of branches $[T]$ is a $\kappa$-compact subspace of $2^{\kappa}$. (Proof: let $\langle \sigma_\alpha \mid \alpha < \kappa\rangle$ be sequence of elements in $2^{<\kappa}$ such that $\{[\sigma_\alpha] \mid \alpha < \kappa\}$ is an open cover of $[T]$. Assume that for every $\alpha < \kappa$ there is $b\in [T]$ such that $b\notin \bigcup_{\beta < \alpha} [\sigma_\beta]$. Then, there is also $t\in T$ with the same property. Let $j\colon \langle V_\kappa, \in, T\rangle\to \langle M, \in \tilde{T}\rangle$ be a weakly compact embedding. By elementarity, there is $t\in \tilde{T}$ such that $t\notin \bigcup_{\alpha < \kappa} [\sigma_\alpha]$, but for all $\alpha < \kappa$, the length of $\sigma_\alpha$ is less than $\kappa$ so we may take $t\in \tilde{T}_\kappa \subseteq [T]$ - a contradiction).

It is consistent that $|[T]| = \kappa^{+} < 2^\kappa$ for a $T\subseteq 2^{<\kappa}$, $\kappa$ weakly compact. For example, let $T$ be the full binary $\kappa$-tree in $L$, and assume that $\kappa^{+} = (\kappa^{+})^L$ but $2^\kappa > \kappa^{+}$. By weakly compactness, every branch of $T$ belongs to $L$. In fact, similar arguments seems to show that if $\kappa$ is weakly compact and there is no tree with exactly $\kappa^{+}$ many branches then there is an inner model with a Woodin cardinal.

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  • $\begingroup$ That's surprising, and important. Unfortunately this does not help for our specific goal, and the problem becomes whether there are uncountable cardinal numbers where the answer is positive. $\endgroup$ – Boaz Tsaban Apr 15 '16 at 10:14
  • $\begingroup$ @BoazTsaban: Do you know if there is a $\kappa$-compact subspace of $2^{\kappa}$ which is not the branches of some subtree? $\endgroup$ – Yair Hayut Apr 15 '16 at 12:39
  • $\begingroup$ @YairHayut: First of all, nice answer -- very enlightening! Secondly, if $\kappa$ is weakly compact, then every $\kappa$-compact subspace of $2^\kappa$ can be represented as $[T]$ for some subtree $T$ of $2^{<\kappa}$. To see this, first notice that "$\kappa$-compact" is equivalent to "closed" in this setting (Lemma 3 from my post). Then, given a closed set $C$, define $T$ to be the set of all $s \in 2^{<\kappa}$ such that $C \cap [s] \neq \emptyset$. Using the fact that $C$ is closed, it's not hard to check that $[T] = C$. $\endgroup$ – Will Brian Apr 15 '16 at 13:32

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