If$p\in X$ define$t(p, X)$ as the least infinite cardinal$\kappa$ such that if$p\in \overline{Y}$ then$p\in \overline{A}$ for some$A\subseteq Y$ with$|A| \leq \kappa$, and let$t(X) := \sup_{p\in X} t(p, X)$ be tightness of$X$. Those are standard cardinal functions in topology.
If$X = \beta \omega\setminus \omega$ then in the articleAn introduction to$\beta\omega$ by van Mill of theHandbook of set-theoretic topology by Kunen and Vaughan, one defines$p\in X$ to be an$R$-point if there exists a cozero set$U\subseteq X$ which witnesses that$t(X) = \mathfrak{c}$, namely we have$p\in\overline{U}$ and for any$A\subseteq U$ with$|A| < \mathfrak{c}$ one has$p\notin\overline{A}$. In lemma 3.3.4 they show$R$-points exist. Hence$t(X)\geq \mathfrak{c}$, while$t(X)\leq s(X)\leq w(X) \leq w(\beta\omega)\leq \mathfrak{c}$ where$t(X)\leq s(X)$ holds for compact Hausdorff$X$, and$w(\beta\omega)\leq \mathfrak{c}$ by the fact that$\{\overline{A} : A\subseteq \omega\}$ is a basis for$\beta\omega$. The inequality$s(X)\leq w(X)$ is easily seen from$w(Z) = |Z|$ for discrete$Z$ and monotonicity of weight$Z\mapsto w(Z)$. This of course implies$t(X)\leq t(\beta \omega)\leq w(\beta \omega)$ and so$t(\beta\omega) = \mathfrak{c}$, since$Z\mapsto t(Z)$ is monotone.
Q. Can we show that if$X\subseteq \beta\omega$ is theNovak space, then$t(X) = \mathfrak{c}$? Or perhaps$t(X) > \omega$?
Note there is no single "Novak space" so this might depend on the particular choice in the construction. It also seems one cannot use lemma 3.3.4 to prove this, since it explicitly uses compactness of$\beta\omega$ to find an$R$-point.
- $\begingroup$@Ulli yes $X\setminus \omega$ will be dense in $\omega^\ast$ since sets of the form $\overline{A}\setminus A$ for infinite $A\subseteq \omega$ form a basis for $\omega^\ast$$\endgroup$Jakobian– Jakobian2025-12-16 16:32:42 +00:00Commented8 hours ago
1 Answer1
When you look at the proof of 3.3.4 in Van Mill's paper that not only does the family$\mathcal{F}$ have the finite intersection property, it does so in a strong sense: if$\mathcal{G}$ is a finite subfamily then$\bigcap\mathcal{G}$ intersects every$C_i$ with$i\ge|\mathcal{G}|$ (there is a typo in the fifth line,$i\le n-1$ should be$i\ge n$).
This implies that$\bigcap\{\overline{F}:F\in\mathcal{F}\}$ is infinite, so that in the construction given onthe$\pi$-base website there will be$\xi$ such that$S_\xi$ is a subset of that intersection, and hence that both$x_\xi$ and$y_\xi$ will be$R$-points. Therefore the subspace$X$, and its sister$Y$, both have tightness$\mathfrak{c}$.
- $\begingroup$To see that $S = \bigcap\{\overline{F} : F\in\mathcal{F}\}$ is infinite, one can partition $\omega$ into infinite sets $I_n$, then $H_n = \bigcup_{i\in I_n} C_i$ will have the property that $S\cap \overline{H_n}\neq\emptyset$ and since $\beta\omega\setminus \omega$ is an $F$-space, we have $\overline{H_n}\cap \overline{H_m} = \emptyset$ for $n\neq m$.$\endgroup$Jakobian– Jakobian2025-12-16 17:20:43 +00:00Commented7 hours ago
- $\begingroup$To see that $p = x_\xi$ being an $R$-point implies $t(p, X) = \mathfrak{c}$ (or at least $t(p, X)\geq \mathfrak{c}$ and one can argue as above that $t(p, X)\leq t(p, \beta\omega)\leq t(\beta\omega)\leq w(\beta\omega) \leq \mathfrak{c}$), note that if $U\subseteq \beta\omega\setminus \omega$ is a cozero set in definition of $R$-point, then $p\in \text{cl}_{X\setminus \omega} (X\cap U)$ from density of $X\setminus \omega$ in $\beta\omega\setminus \omega$, and for any subset $D\subseteq X\cap U$ with $|D|<\mathfrak{c}$ we have $p\notin \text{cl}_{X\setminus \omega} D$.$\endgroup$Jakobian– Jakobian2025-12-16 17:27:27 +00:00Commented7 hours ago
You mustlog in to answer this question.
Explore related questions
See similar questions with these tags.