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":



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> 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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A question about the Stone–Čech compactification of discrete spaces
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. 



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



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


