Let $D(\kappa)$ be the discrete space of cardinality $\kappa$, and $\beta D(\kappa)$ its Stone–Čech compactification.

Is there, for every infinite cardinal $\kappa$, a subset $Y \in [\beta D(\kappa)]^{\kappa^+}$ such that $\psi (Y) = \kappa^+$? Can this be proved in ZFC?

For $\kappa = \aleph_0$, this is true, as a consequence of the Corollary 4.4.5 in the Chapter 11 of the "Handbook of Set-theoretic Topology":

4.4.5. COROLLARY. There is a point $x \in \omega^*$ and a (relatively) discrete sequence $\{ x_{\alpha} : \alpha < \omega_1 \} \subseteq \omega^* \setminus \{ x \}$, such that each neighborhood of $x$ contains all but countably many of the $x_{\alpha}$'s.

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    $\begingroup$ What do you mean by $\psi(Y)$? Is that some sort of cardinal invariant? $\endgroup$ Aug 27, 2013 at 3:03
  • $\begingroup$ Yes, it is the cardinal function "pseudocharacter": encyclopediaofmath.org/index.php/Cardinal_characteristic $\endgroup$ Aug 27, 2013 at 3:33
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    $\begingroup$ First, for each $x \in X$, $\psi (x,X)$ is the smallest number of open sets whose intersection is $\{ x \}$. Then, $\psi (X) = \sup \{ \psi (x,X) : x \in X \}$. $\endgroup$ Aug 27, 2013 at 3:45

2 Answers 2


Here´s a ZFC answer to your question.

All spaces are assumed to be Hausdorff. Recall that the tightness $t(x,X)$ of a point $x$ in the space $X$ is defined as the least cardinal $\kappa$ such that for every set $A \subset X$ such that $x \in \overline{A}$ there exists a $\kappa$-sized subset $B$ of $A$ such that $x \in \overline{B}$. The tightness $t(X)$ of the space $X$ is then defined by taking the supremum of the local tightness over all points $x \in X$.

Arhangel´skii proved that the tightness of a compact space can be characterized by means of a special kind of discrete set called "free sequence". Although we´re not gonna need to use its definition directly, let me recall you what it is.

A set $F \subset X$ is called a free sequence of length $\kappa$ if there exists a well-ordering $\{x_\alpha: \alpha <\kappa \}$ of $F$ such that $\overline{\{x_\alpha: \alpha < \beta \}} \cap \overline{\{x_\alpha: \alpha \geq \beta \}}=\emptyset$, for every $\beta < \kappa$. The freeness $F(X)$ of a space $X$ is the supremum of the cardinalities of its free sequences.

LEMMA 1 (Arhangel´skii) Let $X$ be a compact space. Then $F(x)=t(X)$.

LEMMA 2 (Juhasz and Szentmiklossy) Let $X$ be compact space. If $X$ contains a free sequence of size $\kappa$, then it contains also one that is convergent.

Now, the space $\beta D(\kappa)$ has tightness greater than $\kappa$ (see below). Hence it contains a free sequence of size $\kappa^+$. So it contains a free sequence $F$ of size $\kappa^+$ which converges to some point $x$. Convergence implies that $x$ has pseudocharacter at least $\kappa^+$ in $Y=F \cup \{x\}$. On the other hand, the pseudocharacter of $Y$ cannot exceed $\kappa^+$, because $|Y|=\kappa^+$.

It remains to prove that:

THEOREM: $t(\beta D(\kappa)) \geq \kappa^+$.

There might be a simpler way of proving this, but I could only think of this off the top of my head.

Proof: By a theorem of Nogura ("Tightness of compact Hausdorff spaces and normality of products", J. Math. Soc. Japan 28, No. 2, 1976), the tightness of a space $X$ does not exceed $\kappa$ if and only if $X \times \kappa^+$ is normal, where $\kappa^+$ is provided with its natural order topology. So we just need to prove that $\beta D(\kappa) \times \kappa^+$ is not normal.

To that aim, note first that $(\kappa^++1) \times \kappa^+$ is not normal (you can prove this directly or just use Nogura's theorem cited above). Now $\kappa^++1 \subset 2^{\kappa^+}$ and this last space has density $\kappa$, by the Hewitt-Marczewski-Pondiczery theorem. Let $f$ be any (necessarily continuous) bijection from $D(\kappa)$ to a dense subset $D$ of $2^{\kappa^+}$ such that $|D|=\kappa$. By Corollary 3.6.6 of Engelking's "General Topology" book, $f$ can be extended to a (necessarily perfect) map $F$ from $\beta D(\kappa)$ onto $2^{\kappa^+}$. Since product of perfect maps is perfect we obtain a perfect map from $\beta D(\kappa) \times \kappa^+$ onto $2^{\kappa^+} \times \kappa^+$, but this last space is not normal, because it contains the closed non-normal subspace $(\kappa^++1) \times \kappa^+$. Since normality is preserved by perfect maps, we get that $\beta D(\kappa) \times \kappa^+$ is not normal.

  • $\begingroup$ Good! A reference for the presence of weak $P_\kappa$-points in $\beta D(\kappa)$? $\endgroup$ Mar 18, 2014 at 20:10
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    $\begingroup$ Hi Alberto, I actually meant "weak $P_{\kappa^+}$-point". I don't know whether weak $P_{\kappa^+}$-points exist in $\beta D(\kappa)$ for every cardinal $\kappa$. This appears to have been unknown at least until 2001, or Baker and Kunen would have mentioned it in their paper on Transactions of the AMS. However, I found an alternative proof that $\beta D(\kappa)$ has tightness greater than $\kappa$. See my edit. $\endgroup$ Jul 21, 2014 at 21:38
  • $\begingroup$ Very nice, Santi!! $\endgroup$ Jul 25, 2014 at 3:01

Under $GCH$ the function $\psi$ reflects all infinite cardinals for the class of compact Hausdorff spaces (see for example Theorem 3.8 in "Reflection theorems for cardinal functions" by Hodel and Vaughan). This means that:

if $X$ is a compact Hausdorff space, $\kappa$ is an infinite cardinal and $\psi(X) \geq \kappa$, then there is a $Y \subseteq X$ with $|Y| \leq \kappa$ and $\psi(Y) \geq \kappa$.

Note that $\psi(\beta D(\kappa))=\chi(\beta D(\kappa))=2^\kappa \geq\kappa^+$, so under $GCH$ you get what you want.

I believe it is open whether you need $GCH$ for the quoted reflection theorem, but it should be easier to establish it in this particular context (although I don´t quite see how).

The closest thing I´ve seen resembling Corollary 4.4.5 that you mention, is Theorem 4.5 in Joni Baker´s PhD Thesis (here $u(\kappa)$ denotes the space of uniform ultrafilters):

If $\kappa$ is regular, then there is a set $Y=\lbrace x_\xi : \xi < \kappa^+ \rbrace$ of weak $P_{\kappa^+}$-points of $u(\kappa)$, such that each $x_\xi$ is in the closure of $\lbrace x_\eta : \eta < \kappa^+ \rbrace \setminus \lbrace x_\xi \rbrace$.

Perhaps you can show that $\psi(Y)=\kappa^+$ for that $Y$.


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