A note on the tightness of $G_\delta$-modifications

TL;DR

This paper constructs a specific topological space demonstrating the tightness of its $G_\delta$-modification can exceed certain bounds, addressing a question in the field and exploring the influence of set-theoretic assumptions.

Contribution

It provides a counterexample to a question about the tightness of $G_\delta$-modifications and analyzes how set-theoretic assumptions affect this property.

Findings

Constructed a normal countably tight $T_1$ space with $t(X_\delta) > 2^\omega$.

Showed that under certain set-theoretic conditions, the tightness can be arbitrarily large up to the least $\omega_1$-strongly compact cardinal.

Abstract

We construct a normal countably tight $T_1$ space $X$ with $t(X_\delta) >2^\omega$. This is an answer to the question posed by Dow-Juh\'asz-Soukup-Szentmikl\'ossy-Weiss. We also show that if the continuum is not so large, then the tightness of $G_\delta$-modifications of countably tight spaces can be arbitrary large up to the least $\omega_1$-strongly compact cardinal.

Equations

$$\text{the least ω1-strongly compact}=\sup \{L(X_{\delta })\mid \text{X is compact T2}\}.$$

$$\text{the least ω1-strongly compact}=\sup \{t(X_{\delta })\mid \text{X is normal countably tight T1}\}.$$

Full text

1. Introduction

For a topological space $X$, let $X_{\delta}$ be the $G_{\delta}$-modification of $X$, that is, $X_{\delta}$ is the space $X$ equipped with topology generated by all $G_{\delta}$-subsets of $X$.

Bella and Spadaro [1] studied the connection between the values of various cardinal functions taken on $X$ and $X_{\delta}$, respectively. In their paper they posed the following question: Is $t(X_{\delta})\leq 2^{t(X)}$ true for every (compact) $T_{2}$ space $X$? Recall that $t(X)$, the tightness number of $X$, is the least infinite cardinal $\kappa$ such that for every $A\subseteq X$ and $p\in\overline{A}$, there is $B\in[A]^{\leq\kappa}$ with $p\in\overline{B}$. If $t(X)=\omega$, $X$ is said to be countably tight.

For this question, Dow-Juhász-Soukup-Szentmiklóssy-Weiss [5] answered as follows:

Fact 1.1** ([5]).**

1. (1)

If $X$ is a regular Lindelöf $T_{1}$ space, then $t(X_{\delta})\leq 2^{t(X)}$. 2. (2)

Under $V=L$, for every cardinal $\kappa$ there is a Fréchet-Urysohn space $X$ with $t(X_{\delta})\geq\kappa$.

The clause (1) of Fact 1.1 is a theorem of ZFC. However (2) is a consistency result, and they asked the following natural question:

Question 1.2**.**

Is there a ZFC example of a countably tight Hausdorff (or regular, or Tychonoff) space $X$ for which $t(X_{\delta})>2^{t(X)}$?

In this paper we give a positive answer to their question.

Theorem 1.3**.**

There is a normal countably tight $T_{1}$ space $X$ such that $t(X_{\delta})>2^{\omega}$.

We also observe some connection between $\omega_{1}$-strongly compact cardinals and the tightness of $G_{\delta}$-modifications. Usuba [11] studied the Lindelöf number of $G_{\delta}$-modifications of compact spaces, and proved the following equality:

[TABLE]

Under some assumption, we prove similar results for the tightness of $G_{\delta}$-modifications.

Theorem 1.4**.**

1. (1)

Suppose $\kappa$ is the least $\omega_{1}$-strongly compact cardinal. Then for every countably tight space $X$ we have $t(X_{\delta})\leq\kappa$. 2. (2)

*Suppose there is no weakly Mahlo cardinal $<2^{\omega}$ (e.g., CH holds). If there is no $\omega_{1}$-strongly compact cardinal below a cardinal $\nu$, then there is a normal countably tight $T_{1}$ space $X$ with $t(X_{\delta})\geq\nu$. *

Thus, assuming that $2^{\omega}$ is not so large, we have the following equality:

[TABLE]

Here we present some notations, definitions, and facts.

For a topological space $X$ and $A\subseteq X$, let $\overline{A}^{\delta}$ be the closure of $A$ in $X_{\delta}$.

For a filter $F$ over the set $S$ and a cardinal $\kappa$, let us say that $F$ is $\kappa$-complete if for every family $\mathcal{A}\subseteq F$ of size $<\kappa$, we have $\bigcap\mathcal{A}\in F$. A filter $F$ is $\kappa$-incomplete if $F$ is not $\kappa$-complete.

The concept of $\omega_{1}$-strongly compact cardinal is introduced by Bagaria and Magidor.

Definition 1.5** (Bagaria-Magidor [3, 4]).**

An uncountable cardinal $\kappa$ is $\omega_{1}$-strongly compact if for every set $S$ and every $\kappa$-complete filter $F$ over $S$, the filter $F$ can be extended to an $\omega_{1}$-complete ultrafilter over $S$.

Note that if $\kappa$ is $\omega_{1}$-strongly compact, then every cardinal greater than $\kappa$ is $\omega_{1}$-strongly compact.

Definition 1.6**.**

1. (1)

For an uncountable cardinal $\kappa$ and a set $A$, let $\mathcal{P}_{\kappa}A=\{x\subseteq A\mid\left|{x}\right|<\kappa\}$. 2. (2)

A filter $F$ over $\mathcal{P}_{\kappa}A$ is fine if for every $a\in A$, we have $\{x\in\mathcal{P}_{\kappa}A\mid a\in x\}\in F$.

Fact 1.7** (Bagaria-Magidor [3, 4]).**

1. (1)

An uncountable cardinal $\kappa$ is $\omega_{1}$-strongly compact if and only if for every cardinal $\lambda\geq\kappa$, there exists an $\omega_{1}$-complete fine ultrafilter over $\mathcal{P}_{\kappa}\lambda$. 2. (2)

If $\kappa$ is the least $\omega_{1}$-strongly compact, then $\kappa$ is a limit cardinal and there exists a measurable cardinal $\leq\kappa$. 3. (3)

It is possible that the least $\omega_{1}$-strongly compact is a singular cardinal.

Now we give the proof of (1) in Theorem 1.4. The proof is essentially the same to in Dow-Juhász-Soukup-Szentmiklóssy-Weiss [5], but we give it for the completeness.

Proposition 1.8**.**

Let $\kappa$ be the least $\omega_{1}$-strongly compact. Then for every countably tight topological space $X$, $A\subseteq X$, and $p\in\overline{A}^{\delta}$, there is $B\subseteq A$ with $\left|{B}\right|<\kappa$ and $p\in\overline{B}^{\delta}$. Hence $t(X_{\delta})\leq\kappa$.

Proof.

We may assume that $A=\lambda$ for some cardinal $\lambda\geq\kappa$. By Fact 1.7, there is an $\omega_{1}$-complete fine ultrafilter $U$ over $\mathcal{P}_{\kappa}\lambda$.

Suppose to the contrary that $p\notin\overline{B}^{\delta}$ for every $B\subseteq\lambda$ with $\left|{B}\right|<\kappa$. For $B\subseteq\lambda$ with $\left|{B}\right|<\kappa$, there are open neighborhoods $O^{n}_{B}$ ($n<\omega$) of $p$ with $B\cap\bigcap_{n<\omega}O^{n}_{B}=\emptyset$. Since $U$ is $\omega_{1}$-complete, for each $\alpha<\lambda$, there is $n<\omega$ with $\{B\in\mathcal{P}_{\kappa}\lambda\mid\alpha\in B\setminus\bigcap_{i\leq n}O_{B}^{i}\}\in U$. For $n<\omega$, let $A_{n}$ be the set of all $\alpha<\lambda$ with $\{B\in\mathcal{P}_{\kappa}\lambda\mid\alpha\in B\setminus\bigcap_{i\leq n}O_{B}^{i}\}\in U$. We have $\lambda=\bigcup_{n<\omega}A_{n}$. On the other hand we have $p\notin\overline{A_{n}}$; If $p\in\overline{A_{n}}$, there is a countable $C\subseteq A_{n}$ with $p\in\overline{C}$. Since $U$ is $\omega_{1}$-complete, we can find $B\in\mathcal{P}_{\kappa}\lambda$ with $\alpha\in B\setminus\bigcap_{i\leq n}O_{B}^{i}$ for every $\alpha\in C$, this means that $p\in\bigcap_{i\leq n}O_{B}^{i}$ but $C\cap\bigcap_{i\leq n}O_{B}^{i}=\emptyset$, this is impossible. Thus $p\notin\overline{A_{n}}$, and this immediately implies that $p\notin\overline{A}^{\delta}$. ∎

2. Construction of the spaces

For the sake of constructing our spaces, we use the function spaces. Let us recall some definitions and basic facts. For a Tychonoff space $X$, let $C(X)$ be the set of all continuous functions from $X$ into the real line $\mathbb{R}$. $C_{p}(X)$ is the space $C(X)$ endowed with the point-wise convergence, that is, the topology of $C_{p}(X)$ is generated by the family $\{V(x_{0},\dotsc,x_{n},O_{0},\dotsc,O_{n})\mid x_{0},\dotsc,x_{n}\in X,O_{0},\dotsc,O_{n}\subseteq\mathbb{R}$ are open in $\mathbb{R}\}$ where $V(x_{0},\dotsc,x_{n},O_{0},\dotsc,O_{n})$ is the set of all $f\in C(X)$ with $f(x_{i})\in O_{i}$ for every $i\leq n$.

Fact 2.1** (Arhangel’skiĭ-Pytkeev [2, 9]).**

Let $X$ be a Tychonoff space, and $\nu$ a cardinal. Then $L(X^{n})\leq\nu$ for every $n<\omega$ if and only if $t(C_{p}(X))\leq\nu$. In particular, each finite product of $X$ is Lindelöf if and only if $C_{p}(X)$ is countably tight.

Proposition 2.2**.**

Let $\kappa$ be an uncountable cardinal and $\lambda\geq\kappa$ a cardinal. Suppose there is no $\omega_{1}$-complete fine ultrafilter over $\mathcal{P}_{\kappa}\lambda$. In addition we suppose that, for every countable family $\{U_{n}\mid n<\omega\}$ of fine ultrafilters over $\mathcal{P}_{\kappa}\lambda$, there is a countable partition $\mathcal{A}$ of $\mathcal{P}_{\kappa}\lambda$ such that $A\notin U_{n}$ for every $A\in\mathcal{A}$ and $n<\omega$. Then there is a countably tight Tychonoff space $X$ with $t(X_{\delta})\geq\kappa$.

Proof.

Identifying $\mathcal{P}_{\kappa}\lambda$ as a discrete space, let $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ be the closed subspace of the Stone-Čech compactification $\beta(\mathcal{P}_{\kappa}\lambda)$ consisting of all fine ultrafilters over $\mathcal{P}_{\kappa}\lambda$. Let $X=C_{p}(\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda))$. Since $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ is compact Hausdorff, each finite product of $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ is compact. Hence $X$ is countably tight by Fact 2.1. We shall show that $t(X_{\delta})\geq\kappa$.

Let $\{\mathcal{A}^{\alpha}\mid\alpha<\nu\}$ be an enumeration of all countable partitions of $\mathcal{P}_{\kappa}\lambda$. For $\alpha<\nu$, let $S^{\mathcal{A}^{\alpha}}=\{U\in\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)\mid A\notin U$ for every $A\in\mathcal{A}^{\alpha}\}$. $S^{\mathcal{A}^{\alpha}}$ is a closed $G_{\delta}$-subset of $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$. Since there is no $\omega_{1}$-complete fine ultrafilter over $\mathcal{P}_{\kappa}\lambda$, the family $\{S^{\mathcal{A}^{\alpha}}\mid\alpha<\nu\}$ is a cover of $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$. Furthermore, by our assumption, for every $U_{n}\in\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ ($n<\omega$) there is $\alpha<\nu$ such that $U_{n}\in S^{\mathcal{A}^{\alpha}}$ for every $n<\omega$.

We use the following fact:

Fact 2.3** (Usuba [11]).**

If $\{S^{\mathcal{A}^{\alpha}}\mid\alpha\in E\}$ is a cover of $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ for some $E\subseteq\nu$, then $\left|{E}\right|\geq\kappa$.

Sketch of the proof.

Take $E\subseteq\nu$ with size $<\kappa$. Take a sufficiently large regular $\theta$, and take $M\prec H_{\theta}$ containing all relevant objects such that $\left|{M}\right|<\kappa$ and $E\subseteq M$. For each $\alpha\in E$, there is a unique $A_{\alpha}\in\mathcal{A}^{\alpha}$ with $M\cap\kappa\in A_{\alpha}$. By the elementarity, we have that for every $\alpha_{0},\dotsc,\alpha_{n}\in E$ and $\beta_{0},\dotsc,\beta_{m}<\lambda$, the family $\{x\in\mathcal{P}_{\kappa}\lambda\mid x\in A_{\alpha_{i}}$ for every $i<n$ and $\beta_{i}\in x$ for every $i<m\}$ is non-empty. Thus we can take a fine ultrafilter $U$ over $\mathcal{P}_{\kappa}\lambda$ such that $U\notin S^{\mathcal{A}^{\alpha}}$ for every $\alpha\in E$. Hence $\{S^{\mathcal{A}^{\alpha}}\mid\alpha\in E\}$ is not a cover. ∎

For $\alpha<\nu$, since $S^{\mathcal{A}^{\alpha}}$ is a closed $G_{\delta}$-subset of the compact Hausdorff space $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$, there is a continuous map $f_{\alpha}:\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)\to[0,1]$ such that $S^{\mathcal{A}^{\alpha}}=f_{\alpha}^{-1}\{0\}$. Let $D=\{f_{\alpha}\mid\alpha<\nu\}$. Let $g:\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)\to\{0\}$ be the constant function. We shall show that $g$ and $D$ witness $t(X_{\delta})\geq\kappa$.

First we check that $g\in\overline{D}^{\delta}$. Take an open neighborhood $O$ of $g$ in $X_{\delta}$. By the definition of the topology of $X_{\delta}$, there are $U_{n}\in\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ $(n<\omega)$ such that $\{h\in X\mid h(U_{n})=0$ for $n<\omega\}\subseteq O$. By the assumption, there is $\alpha<\nu$ such that $U_{n}\in S^{\mathcal{A}^{\alpha}}$ for every $n<\omega$. Then $f_{\alpha}(U_{n})=0$ for $n<\omega$, hence $f_{\alpha}\in D\cap O$.

Finally we show that if a set $E\subseteq\nu$ has cardinality $<\kappa$, then $g\notin\overline{\{f_{\alpha}\mid\alpha\in E\}}^{\delta}$, which means that $t(X_{\delta})\geq\kappa$. Suppose to the contrary that $g\in\overline{\{f_{\alpha}\mid\alpha\in E\}}^{\delta}$. For each $U\in\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$, the set $\{h\in C_{p}(X)\mid h(U)=0\}$ is an open neighborhood of $g$ in $X_{\delta}$. So we can pick $\alpha\in E$ with $f_{\alpha}(U)=0$. Since $S^{\mathcal{A}^{\alpha}}=f_{\alpha}^{-1}\{0\}$, we have $U\in S^{\mathcal{A}^{\alpha}}$. This shows that $\{S^{\mathcal{A}^{\alpha}}\mid\alpha\in E\}$ is a cover of $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$, contradicting to Fact 2.3. ∎

3. On the assumptions of Proposition 2.2

Now let us discuss when the assumptions of Proposition 2.2 hold.

For a filter $F$ over the set $S$, let $F^{+}=\{X\in\mathcal{P}(S)\mid S\setminus X\notin F\}$. $F^{+}$ is the complement of the dual ideal of $F$. An element of $F^{+}$ is called an $F$-positive set. For $X\in F^{+}$, let $F\restriction X=\{Y\subseteq S\mid Y\cup(S\setminus X)\in F\}$. $F\restriction X$ is the filter over $S$ generated by $F\cup\{X\}$.

Lemma 3.1**.**

Let $S$ be an uncountable set, and $\{U_{n}\mid n<\omega\}$ a family of ultrafilters over $S$. Let $F=\bigcap_{n<\omega}U_{n}$.Then the following are equivalent:

1. (1)

*There is a countable partition $\mathcal{A}$ of $S$ such that $A\notin U_{n}$ for every $A\in\mathcal{A}$ and $n<\omega$. * 2. (2)

For every $X\in F^{+}$, the filter $F\restriction X$ is $\omega_{1}$-incomplete.

Proof.

(1) $\Rightarrow$ (2) is clear. For (2) $\Rightarrow$ (1), we define $C_{\alpha}\subseteq S$ $(\alpha<\omega_{1})$ as follows: First, let $C_{0}=S\in F$. Suppose $C_{\gamma}$ is defined for every $\gamma<\alpha$ so that:

1. (1)

$\langle{C_{\gamma}\mid\gamma<\alpha}\rangle$ is a $\subseteq$-decreasing sequence of $F$-positive sets. 2. (2)

$C_{\gamma}=\bigcap_{\delta<\gamma}C_{\delta}$ if $\gamma$ is limit. 3. (3)

If $\gamma+1<\alpha$, then there are $C_{\gamma,i}\in F\restriction C_{\gamma}$ for $i<\omega$ such that $C_{\gamma}=C_{\gamma,0}\supseteq C_{\gamma,1}\supseteq\cdots$ and $C_{\gamma+1}=\bigcap_{i<\omega}C_{\gamma,i}\notin F\restriction C_{\gamma}$.

Suppose $\alpha=\beta+1$. Since $C_{\beta}\in F^{+}$ and $F\restriction C_{\beta}$ is $\omega_{1}$-incomplete, we can find $C_{\beta,i}\subseteq C_{\beta}$ for $i<\omega$ such that $C_{\beta}=C_{\beta,0}\supseteq C_{\beta,1}\supseteq\cdots$, $C_{\beta,i}\in F\restriction C_{\beta}$ for every $i<\omega$, and $\bigcap_{i<\omega}C_{\beta,i}\notin F\restriction C_{\beta}$. If $\bigcap_{i<\omega}C_{\beta,i}\notin F^{+}$, then we finish this construction. If $\bigcap_{i<\omega}C_{\beta,i}\in F^{+}$, then let $C_{\alpha}=\bigcap_{i<\omega}C_{\beta,i}$.

If $\alpha$ is limit and $\bigcap_{\gamma<\alpha}C_{\gamma}\notin F^{+}$, then we finish the construction, otherwise let $C_{\alpha}=\bigcap_{\gamma<\alpha}C_{\gamma}$.

We claim that this construction have to be finished at some $\gamma<\omega_{1}$. If not, then let $D_{\alpha}=C_{\alpha}\setminus C_{\alpha+1}$ for $\alpha<\omega_{1}$. Since $C_{\alpha+1}\notin F\restriction C_{\alpha}$, we have $D_{\alpha}\in F^{+}$. For $\alpha<\omega_{1}$, there is some $n_{\alpha}<\omega$ with $D_{\alpha}\in U_{n_{\alpha}}$. Hence there is some $n<\omega$ such that the set $\{\alpha<\omega_{1}\mid n_{\alpha}=n\}$ is uncountable. Pick $\alpha<\beta<\omega_{1}$ with $n=n_{\alpha}=n_{\beta}$. We have $D_{\alpha},D_{\beta}\in U_{n}$, hence $D_{\alpha}\cap D_{\beta}\neq\emptyset$. This is impossible.

Now suppose $\{C_{\alpha}\mid\alpha<\gamma\}$ is defined as above but $C_{\gamma}$ cannot be defined. If $\gamma$ is limit, we have $\bigcap_{\alpha<\gamma}C_{\alpha}\notin F^{+}$. By shrinking each $C_{\alpha}$, we may assume that $\bigcap_{\alpha<\gamma}C_{\alpha}=\emptyset$. By the construction, for $\alpha<\gamma$, there is a sequence $\langle{C_{\alpha,i}\mid i<\omega}\rangle$ as above. Let $A_{\alpha,i}=C_{\alpha,i}\setminus C_{\alpha,i+1}$. We check that $A_{\alpha,i}\notin F^{+}$; Since $C_{\alpha,i},C_{\alpha,i+1}\in F\restriction C_{\alpha}$, we have that $A_{\alpha,i}=C_{\alpha,i}\setminus C_{\alpha,i+1}\notin(F\restriction C_{\alpha})^{+}$. Furthermore, since $A_{\alpha,i}\subseteq C_{\alpha,i}\subseteq C_{\alpha}$, we have $A_{\alpha,i}\notin F^{+}$. Now the family $\{A_{\alpha,i}\mid\alpha<\gamma,i<\omega\}$ is a countable partition of $S$ such that $A_{\alpha,i}\notin F^{+}$, so $A_{\alpha,i}\notin U_{n}$ for every $n<\omega$. Thus (1) holds.

If $\beta$ is successor, say $\gamma=\beta+1$, then there are $C_{\beta,i}\in(F\restriction C_{\beta})^{*}$ ($i<\omega$) such that $C_{\beta}\supseteq C_{\beta,0}\supseteq C_{\beta,1}\supseteq\cdots$ and $\bigcap_{i<\omega}C_{\beta,i}\notin F^{+}$. As in the limit case, we may assume that $\bigcap_{i<\omega}C_{\beta,i}=\emptyset$. For each $\alpha\leq\beta$, let $A_{\alpha,i}=C_{\alpha,i}\setminus C_{\alpha,i+1}$. Then $\{A_{\alpha,i}\mid i<\omega,\alpha\leq\beta\}$ is a required partition. ∎

We will use the generic ultrapower argument. See Foreman [6] for the generic ultrapower. Here we present some basic definitions and facts.

Let $F$ be a filter over the set $S$. For a cardinal $\kappa$, we say that $F$ is $\kappa$-saturated if for every family $\{X_{\alpha}\mid\alpha<\kappa\}$ of $F$-positive sets, there are $\alpha<\beta<\kappa$ with $X_{\alpha}\cap X_{\beta}\in F^{+}$. For $F$-positive sets $X$ and $Y$, define $X\leq_{F}Y$ if $X\setminus Y\notin F^{+}$. Let $\mathbb{P}_{F}$ be the poset $F^{+}$ with the order $\leq_{F}$. Note that for $X,Y\in\mathbb{P}_{F}$, $X$ is compatible with $Y$ in $\mathbb{P}_{F}$ if and only if $X\cap Y\in F^{+}$, and $\mathbb{P}_{F}$ has the $\kappa$-c.c. if and only if $F$ is $\kappa$-saturated.

If $G$ is a $(V,\mathbb{P}_{F})$-generic filter, then $G$ is a $V$-ultrafilter over $S$, that is the following hold:

• •

$S\in G$, $\emptyset\notin G$.

• •

$X\cap Y\in G$ for every $X,Y\in G$.

• •

For $X,Y\in V$, if $X\in G$ and $X\subseteq Y\subseteq S$ then $Y\in G$.

• •

For every $X\subseteq S$ with $X\in V$, either $X\in G$ or $S\setminus X\in G$.

Hence we can take the generic ultrapower of $V$ by $G$. For a $(V,\mathbb{P}_{F})$-generic filter $G$, let $\mathrm{Ult}(V,G)$ be the generic ultrapower of $V$ by $G$, and $j:V\to\mathrm{Ult}(V,G)$ be the elementary embedding induced by $G$. If $\mathrm{Ult}(V,G)$ is well-founded, we identify $\mathrm{Ult}(V,G)$ with its transitive collapse. We say that $F$ is precipitous if for every $(V,\mathbb{P}_{F})$-generic $G$, $\mathrm{Ult}(V,G)$ is well-founded.

Fact 3.2**.**

*Let $\kappa$ be an uncountable cardinal, and $F$ a $\kappa$-complete filter over $S$. If $F$ is $\kappa^{+}$-saturated, then $F$ is precipitous. *

Proposition 3.3**.**

Let $\kappa$ be an uncountable cardinal, and $\lambda>2^{\kappa}$. Let $\{U_{n}\mid n<\omega\}$ be a family of $\omega_{1}$-incomplete fine ultrafilters over $\mathcal{P}_{\kappa}\lambda$. If the filter $F=\bigcap_{n<\omega}U_{n}$ is $\omega_{1}$-complete, then there is a weakly Mahlo cardinal $<2^{\omega}$, and $\kappa>(2^{\omega})^{+}$.

Proof.

First note that for $X\subseteq\mathcal{P}_{\kappa}\lambda$, $X\in\mathbb{P}_{F}$ if and only if $X\in U_{n}$ for some $n<\omega$. For each $X\in\mathbb{P}_{F}$, let $I_{X}=\{n<\omega\mid X\in U_{n}\}\neq\emptyset$.

We shall prove a series of claims.

Claim 3.4**.**

For $X,Y\in\mathbb{P}_{F}$, $X\leq_{F}Y\iff I_{X}\subseteq I_{Y}$, and $X$ is compatible with $Y\iff I_{X}\cap I_{Y}\neq\emptyset$.

Proof.

If $I_{X}\nsubseteq I_{Y}$, pick $n\in I_{X}\setminus I_{Y}$. We know $X\in U_{n}$ but $Y\notin U_{n}$, hence $X\setminus Y\in U_{n}$, and $X\not\leq_{F}Y$. For the converse, suppose $X\not\leq_{F}Y$. Then $X\setminus Y\in F^{+}$, and there is $n<\omega$ with $X\setminus Y\in U_{n}$. We have $X\in U_{n}$ but $Y\notin U_{n}$, so $n\in I_{X}\setminus I_{Y}$ and $I_{X}\nsubseteq I_{Y}$.

If $X$ is compatible with $Y$, then there is $Z\leq_{F}X,Y$. We may assume $Z\subseteq X\cap Y$, then $I_{Z}\subseteq I_{X}\cap I_{Y}$, so $I_{X}\cap I_{Y}\neq\emptyset$. For the converse, if $I_{X}\cap I_{Y}\neq\emptyset$, take $n\in I_{X}\cap I_{Y}$. Then $X,Y\in U_{n}$, so $X\cap Y\in U_{n}$. Hence $X\cap Y\in F^{+}$, and we have $X\cap Y\leq_{F}X,Y$. ∎

Claim 3.5**.**

$\mathbb{P}_{F}$* has the c.c.c., and has a dense subset of size $\leq 2^{\omega}$.*

Proof.

Take an uncountable family $\{X_{\alpha}\mid\alpha<\omega_{1}\}\subseteq\mathbb{P}_{F}$. Since $I_{X_{\alpha}}\subseteq\omega$, there must be $\alpha<\beta<\omega_{1}$ with $I_{X_{\alpha}}\cap I_{X_{\beta}}\neq\emptyset$. Hence $X_{\alpha}$ is compatible with $X_{\beta}$ by Claim 3.4.

For $X,Y\in\mathbb{P}_{F}$, define $X\approx Y$ if $X\leq_{F}Y$ and $Y\leq_{F}X$. By Claim 3.4, $X\approx Y$ if and only if $I_{X}=I_{Y}$. Hence there are at most $2^{\omega}$ many equivalence classes, and we can take a dense subset in $\mathbb{P}_{F}$ of size $\leq 2^{\omega}$. ∎

From now on, we identify $\mathbb{P}_{F}$ with its dense subset of size $\leq 2^{\omega}$.

Now, we know that $F$ is $\omega_{1}$-complete and $\omega_{1}$-saturated. Hence $F$ is precipitous by Fact 3.2. Take a $(V,\mathbb{P}_{F})$-generic $G$, and let $j:V\to\mathrm{Ult}(V,G)$ be the generic elementary embedding induced by $G$. Since $F$ is precipitous, we can identify $\mathrm{Ult}(V,G)$ with its transitive collapse $M$. For a map $f:\mathcal{P}_{\kappa}\lambda\to V$ with $f\in V$, let $[f]$ be the equivalence class of $f$ by $G$. If $id:\mathcal{P}_{\kappa}\lambda\to\mathcal{P}_{\kappa}\lambda$ is the identity map, then we have $j``\lambda\subseteq[id]$ because $F$ is a fine filter. Moreover $\lambda\leq\left|{[id]}\right|^{M}<j(\kappa)$.

Let $\mu$ be the critical point of $j$. $\mu$ is regular uncountable in $V$. Since $\mathbb{P}_{F}$ has the c.c.c., $\mu$ remains regular in $V[G]$, and so does in $M$.

Claim 3.6**.**

$\mu$* is weakly Mahlo in $V$.*

Proof.

Let $C\subseteq\mu$ be a club in $\mu$ with $C\in V$. Then $\mu\in j(C)$, hence it holds that the statement “$j(C)$ contains a regular cardinal” in $M$. By the elementarity of $j$, $C$ contains a regular cardinal in $V$. ∎

Note that $\mu$ is in fact weakly $\mu$-Mahlo.

Claim 3.7**.**

For every cardinal $\nu$, we have $(2^{\nu})^{V}=(2^{\nu})^{V[G]}$.

Proof.

Since $\mathbb{P}_{F}$ has a dense subset of size $(2^{\omega})^{V}$ and has the c.c.c., there are at most $(2^{\omega})^{\omega}$-many antichains in $\mathbb{P}_{F}$, and at most $((2^{\omega})^{\nu})^{V}=(2^{\nu})^{V}$ many canonical names for subsets of $\nu$. Hence we have $(2^{\nu})^{V}=(2^{\nu})^{V[G]}$. ∎

Thus for each cardinal $\nu$, we can let $2^{\nu}$ denote $(2^{\nu})^{V}$ and $(2^{\nu})^{V[G]}$. In addition, since $\mathbb{P}_{F}$ has the c.c.c., we have $(\nu^{+})^{V}=(\nu^{+})^{V[G]}$ for every cardinal $\nu$, and we can let $\nu^{+}$ denote $(\nu^{+})^{V}$ and $(\nu^{+})^{V[G]}$. Note that $(\nu^{+})^{M}\leq\nu^{+}$ for every cardinal $\nu$.

Claim 3.8**.**

$j(2^{\omega})<(2^{\omega})^{+}=j((2^{\omega})^{+})<\kappa$.

Proof.

If $j(2^{\omega})\geq(2^{\omega})^{+}$, then $M$ has at least $(2^{\omega})^{+}$ many subsets of $\omega$, so we have $(2^{\omega})^{V[G]}\geq(2^{\omega})^{+}$, this contradicts to the previous claim, and we have $j(2^{\omega})<(2^{\omega})^{+}$. Thus $(2^{\omega})^{+}\leq j((2^{\omega})^{+})=(j(2^{\omega})^{+})^{M}\leq j(2^{\omega})^{+}\leq(2^{\omega})^{+}$, so we have $j((2^{\omega})^{+})=(2^{\omega})^{+}$.

Finally, since $(2^{\omega})^{+}\leq(2^{\kappa})^{+}\leq\lambda$, we have $j((2^{\omega})^{+})=(2^{\omega})^{+}\leq\lambda<j(\kappa)$. Then $(2^{\omega})^{+}<\kappa$ by the elementarity of $j$. ∎

We completes the proof by showing the following:

Claim 3.9**.**

$\mu<2^{\omega}$.

Proof.

Note that $j(\mu)>\mu^{+}$ since $\mathcal{P}(\mu)^{V}\subseteq\mathcal{P}(\mu)^{M}$ and $j(\mu)$ is a limit cardinal in $M$.

Since $j(2^{\omega})<(2^{\omega})^{+}$, we have $\mu\neq 2^{\omega}$. Next we show $2^{\omega}>\mu$. If not, then $2^{\omega}<\mu$. Take $X\in G$ such that $X\Vdash_{\mathbb{P}_{F}}$“the critical point of $j$ is $\mu$”. Then the filter $F\restriction X$ is in fact $\mu$-complete. We use the following well-known fact by Tarski:

Fact 3.10** (Tarski, e.g. see Kanamori [8]).**

Let $S$ be an uncountable set, and $F$ a filter over $S$. If $F$ is $(2^{\omega})^{+}$-complete and $\omega_{1}$-saturated, then there is $Y\in F^{+}$ such that $F\restriction Y$ is an ultrafilter.

Since $2^{\omega}<\mu$, by Tarski’s theorem there is $Y\in(F\restriction X)^{+}$ such that $F\restriction Y$ is an ultrafilter. Hence $F\restriction Y=U_{n}$ for some $n$, and $U_{n}$ is $\mu$-complete. This is a contradiction. ∎

∎

Lemma 3.11**.**

Let $X$ be a countably tight $T_{1}$ space. Then there is a normal countably tight $T_{1}$ space $Y$ with $t(Y_{\delta})=t(X_{\delta})$.

Proof.

Fix a point $p^{*}\in X$ such that there is $A\subseteq X$ with $p^{*}\in\overline{A}^{\delta}$, but no $B\subseteq A$ with $\left|{B}\right|<t(X_{\delta})$ and $p\in\overline{B}^{\delta}$. Let $Y$ be the space $X$ equipped with the following topology:

1. (1)

Every $q\in X\setminus\{p^{*}\}$ is isolated in $Y$. 2. (2)

A local base for $p^{*}$ in $Y$ is the same to in $X$.

It is easy to check that $Y$ is a normal countably tight $T_{1}$ space with $t(Y_{\delta})=t(X_{\delta})$. ∎

Now we have the theorems.

Corollary 3.12**.**

There is a normal countably tight $T_{1}$ space $X$ such that $t(X_{\delta})>2^{\omega}$.

Proof.

Let $\kappa=(2^{\omega})^{+}$. $\kappa$ is not $\omega_{1}$-strongly compact, and so there is $\lambda\geq\kappa$ such that $\mathcal{P}_{\kappa}\lambda$ has no $\omega_{1}$-complete fine ultrafilter. $\lambda$ can be arbitrary large, so we may assume $\lambda>2^{\kappa}$. By Proposition 2.2 and Lemmas 3.1, 3.11, it is enough to show that for every countable family $\{U_{n}\mid n<\omega\}$ of fine ultrafilters over $\mathcal{P}_{\kappa}\lambda$ and $X\in F^{+}=(\bigcap_{n<\omega}U_{n})^{+}$, the filter $F\restriction X$ is $\omega_{1}$-incomplete. Let $I=\{n<\omega\mid X\in U_{n}\}$. Then it is easy to check that $F\restriction X=\bigcap_{n\in I}U_{n}$. Because $\kappa=(2^{\omega})^{+}$, we know that $\bigcap_{n\in I}U_{n}$ is $\omega_{1}$-incomplete by Proposition 3.3. ∎

Corollary 3.13**.**

Suppose there is no weakly Mahlo cardinal $<2^{\omega}$. If there is no $\omega_{1}$-strongly compact cardinal below a cardinal $\kappa$, then there is a normal countably tight $T_{1}$ space $X$ with $t(X_{\delta})\geq\kappa$.

Proof.

We know that $\kappa$ is not $\omega_{1}$-strongly compact, so there is a large $\lambda>\kappa$ such that $\mathcal{P}_{\kappa}\lambda$ cannot carry an $\omega_{1}$-complete fine ultrafilter. By the assumption and Proposition 3.3, there is no countable family of fine ultrafilters $\{U_{n}\mid n<\omega\}$ over $\mathcal{P}_{\kappa}\lambda$ with $\bigcap_{n<\omega}U_{n}$ $\omega_{1}$-complete. Again, by Proposition 2.2 and Lemmas 3.1, 3.11 we can take a normal countably tight $T_{1}$ space $X$ with $t(X_{\delta})\geq\kappa$. ∎

Question 3.14**.**

Is the equality that “the least $\omega_{1}$-strongly compact$=\sup\{t(X_{\delta})\mid X$ is normal countably tight $T_{1}\}$” provable from ZFC without any assumptions?

Question 3.15**.**

In the theorems, can we replace “countably tight” by “Fréchet-Urysohn”? For instance, is there a ZFC-example of a Fréchet-Urysohn space $X$ with $t(X_{\delta})>2^{\omega}$?

Note that our space $C_{p}(\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda))$ is not Fréchet-Urysohn; It is known that for a compact Hausdorff space $Y$, $C_{p}(Y)$ is Fréchet-Urysohn if and only if $Y$ is scattered (Pytkeev [10], Gerlits [7]). However the space $\mathrm{Fine}(\mathcal{P}_{\kappa}\lambda)$ is not scattered.

Acknowledgments

This research was supported by JSPS KAKENHI Grant Nos. 18K03403 and 18K03404.

Index