Proof. Let $a_n\in \mathsf {Fin}$ for $n\in \omega $ be pairwise disjoint. Using Lemma 3.1 together with the assumption $\kappa <\mathfrak {t}$ , one can easily construct $w_\alpha \in \mathcal {F}_\alpha $ and $I_\alpha \in [\omega ]^\omega $ for $\alpha \in \kappa $ such that the following conditions are satisfied:

• • $w_\alpha \cap a_n=\varnothing $ for every $\alpha \in \kappa $ and $n\in I_\alpha $ .

• • $I_\beta \subseteq ^\ast I_\alpha $ whenever $\alpha \leq \beta $ .

Once the construction is completed, let I be a pseudointersection of $\{I_\alpha :\alpha \in \kappa \}$ . This means that we can fix $n_\alpha \in \omega $ for $\alpha \in \kappa $ such that $w_\alpha \cap a_n =\varnothing $ for every $n\in I\setminus n_\alpha $ . Finally, set

$$ \begin{align*}z_\alpha=w_\alpha\setminus\bigcup_{n\in n_\alpha}a_n \end{align*} $$

for $\alpha \in \kappa $ . It is easy to realize that the $z_\alpha $ and I are as desired.

Proof. Start by choosing an arbitrary $r_\varnothing \in Q$ . Fix $v_\alpha \in \mathcal {F}_\alpha $ for $\alpha \in \kappa $ such that $v_\alpha \subseteq ^\ast \pi _\alpha (r)$ for all $r\in Q$ and $\alpha \in \kappa $ . Also assume that each $v_\alpha \subseteq \pi _\alpha (r_\varnothing )$ . Define

$$ \begin{align*}\mathcal{S}=\{\varnothing\}\cup\{s\in 2^{<\omega}:|s|\geq 1\text{ and }s(|s|-1)=0\}. \end{align*} $$

Also set $\mathcal {S}_{\leq n}=\{s\in \mathcal {S} :|s|\leq n\}$ and $\mathcal {S}_n=\{s\in \mathcal {S} :|s|=n\}$ for $n\in \omega $ . We will need a bookkeeping function $\psi :\omega \longrightarrow \omega $ that satisfies the following requirements:

1. (1) $\psi (i)<i$ for every $i\geq 1$ .

2. (2) $\psi ^{-1}(j)$ is infinite for every $j\in \omega $ .

Constructing such a function is an easy exercise, left to the reader.

Denote by $\mathbb {P}$ the set of all quadruples of the form $p=(n,F,\zeta ,\vec {k})=(n^p,F^p,\zeta ^p,\vec {k}^p)$ that satisfy the following requirements:

1. (3) $n\in \omega $ .

2. (4) $F\in [\kappa ]^{<\omega }$ .

3. (5) $\zeta :\mathcal {S}_{\leq n}\longrightarrow Q$ and $\zeta (\varnothing )=r_\varnothing $ .

4. (6) $s\subsetneq s'$ implies $\zeta (s)\neq \zeta (s')$ for all $s,s'\in \mathcal {S}_{\leq n}$ .

5. (7) $\vec {k}=(k_0,\ldots ,k_n)\in \omega ^{n+1}$ , where $0=k_0<\cdots <k_n$ .

We will also use the obvious notation $\vec {k}^p=(k^p_0,\ldots ,k^p_n)$ for $p\in \mathbb {P}$ , where $n=n^p$ . Define $\ell _s\in \omega $ for every $s\in 2^{<\omega }$ so that $\{t\in \mathcal {S}:t\subsetneq s\}=\{t^s_i:i<\ell _s\}$ , where $\varnothing =t^s_0\subsetneq \cdots \subsetneq t^s_{\ell _s-1}$ whenever $\ell _s\geq 1$ . Notice that if $s,s'\in 2^{<\omega }$ and $s\subseteq s'$ then $t^s_i=t^{s'}_i$ for every $i<\ell _s$ .

Order $\mathbb {P}$ by declaring $q\leq p$ if the following conditions are satisfied:

1. (8) $n^q\geq n^p$ .

2. (9) $F^q\supseteq F^p$ .

3. (10) $\zeta ^q\supseteq \zeta ^p$ .

4. (11) $\vec {k}^q$ is an end-extension of $\vec {k}^p$ (i.e., $k^q_i=k^p_i$ for every $i\leq n^p$ ).

5. (12) $\zeta ^q(s)\upharpoonright \big (F^p\times k^q_{|s|-1}\big )=\zeta ^q\big (t^s_{\psi (\ell _s)}\big )\upharpoonright \big (F^p\times k^q_{|s|-1}\big )$ for every $s\in \mathcal {S}$ such that $n^p<|s|\leq n^q$ .

6. (13) $v_\alpha \setminus k^q_n\subseteq \bigcap \big \{\pi _\alpha \big (\zeta ^q(s)\big ):s\in \mathcal {S}_{\leq n}\big \}$ whenever $\alpha \in F^p$ and $n^p<n\leq n^q$ .

Given $\alpha \in \kappa $ and $n\in \omega $ , make the following definitions:

• • $D^{\mathsf {coord}}_\alpha =\{p\in \mathbb {P}:\alpha \in F^p\}$ .

• • $D^{\mathsf {length}}_n=\{p\in \mathbb {P}:n^p\geq n\}$ .

Claim 1. Each $D^{\mathsf {coord}}_\alpha $ is a dense subset of $\mathbb {P}$ .

Proof. This is trivial.

Claim 2. Each $D^{\mathsf {length}}_n$ is a dense subset of $\mathbb {P}$ .

Proof. Pick $n\in \omega $ . Clearly, it will be enough to show that for every $p\in \mathbb {P}$ there exists $q\in \mathbb {P}$ such that $q\leq p$ and $n^q=n^p+1$ . So pick $p=(n,F,\zeta ,\vec {k})\in \mathbb {P}$ . Begin to construct q by setting $n^q=n+1$ and $F^q=F$ . In order to satisfy condition $(10)$ , we must set $\zeta ^q(s)=\zeta (s)$ for $s\in \mathcal {S}_{\leq n}$ . Given $s\in \mathcal {S}_{n+1}$ , using condition $(1)$ and the assumption that Q is crowded, we can define $\zeta ^q(s)$ to be any element of

$$ \begin{align*}\big\{r\in Q:r\upharpoonright (F\times k_n)=\zeta\big(t^s_{\psi(\ell_s)}\big)\upharpoonright (F\times k_n)\big\}\setminus\{\zeta(t^s_i):i<\ell_s\}. \end{align*} $$

This way, conditions $(6)$ and $(12)$ will be satisfied. In order to satisfy $(11)$ , we must set $k^q_i=k^p_i$ for $i\leq n$ . To conclude the construction of q, simply choose $k^q_{n+1}>k^p_n$ big enough so that condition $(13)$ will be satisfied.

Claim 3. $\mathbb {P}$ is $\sigma \text {-}\mathrm {centered}$ .

Proof. Notice that there are only countably many triples $(n,\zeta ,\vec {k})$ such that the requirements $(3)$ and $(5)$ – $(7)$ are satisfied. Given such a triple, set

$$ \begin{align*}\mathbb{P}(n,\zeta,\vec{k})=\{p\in\mathbb{P}:n^p=n\text{, }\zeta^p=\zeta\text{ and }\vec{k}^p=\vec{k}\}. \end{align*} $$

Since each $\mathbb {P}(n,\zeta ,\vec {k})$ is clearly centered, the desired conclusion follows.

Claim 4. $Q'$ is crowded.

Proof. Pick $r\in Q'$ . Let $F\in [\kappa ]^{<\omega }$ and $\ell \in \omega $ . We will find $r'\in Q'$ such that $r'\neq r$ and $r'\upharpoonright (F\times \ell )=r\upharpoonright (F\times \ell )$ . Fix $n\geq 1$ such that $k_{n-1}\geq \ell $ . Using Claims 1 and 2, it is possible to find $p\in G$ such that $n^p\geq n$ and $F^p\supseteq F$ . Let $s\in \mathcal {S}$ be such that $s\subseteq \phi $ and $\zeta (s)=r$ . Let $j\in \omega $ be the unique index such that $s=t^{s'}_j$ for every $s'\in 2^{<\omega }$ such that $s\subsetneq s'$ . By condition $(2)$ , it is possible to find $s'\in \mathcal {S}$ such that $s\subsetneq s'\subseteq \phi $ , $|s'|>n^p$ and $\psi (\ell _{s'})= j$ . Set $r'=\zeta (s')$ , and observe that $r'\neq r$ by condition $(6)$ . Finally, pick $q\in G$ such that $q\leq p$ and $n^q\geq |s'|$ . It follows from condition $(12)$ that

$$ \begin{align*}r'\upharpoonright (F\times k_{n-1})=\zeta\big(t^{s'}_{\psi(\ell_{s'})}\big)\upharpoonright (F\times k_{n-1}). \end{align*} $$

Since $\zeta \big (t^{s'}_{\psi (\ell _{s'})}\big )=\zeta \big (t^{s'}_j\big )=\zeta (s)=r$ , this shows that $r'$ is as desired.

Claim 5. $\pi _\alpha [Q']\subseteq z_\alpha \!\uparrow $ for every $\alpha \in \kappa $ .

Proof. Pick $\alpha \in \kappa $ . We will use induction to show that $z_\alpha \subseteq \pi _\alpha \big (\zeta (\phi \upharpoonright n)\big )$ for every $n\in \omega $ such that $\phi \upharpoonright n\in \mathcal {S}$ . The case $n=0$ is clear by our choice of $v_\alpha $ . Now assume that $n\geq 1$ and $s=\phi \upharpoonright n\in \mathcal {S}$ . If $n\leq n^{p_\alpha }$ then $z_\alpha \subseteq \pi _\alpha \big (\zeta (s)\big )$ by our choice of $n_\alpha $ , so assume that $n>n^{p_\alpha }$ . Let $k\in z_\alpha $ . We will show that $k\in \pi _\alpha \big (\zeta (s)\big )$ by considering the following cases:

• • $k\in [0,k_{n-1})$ .

• • $k\in [k_{n-1},k_n)$ .

• • $k\in [k_n,\infty )$ .

First assume that $k\in [0,k_{n-1})$ . Pick $q\in G$ such that $q\leq p_\alpha $ and $n^q\geq n$ . Since we have assumed that $n>n^{p_\alpha }$ , condition $(12)$ guarantees that

$$ \begin{align*}\zeta(s)\upharpoonright (\{\alpha\}\times k_{n-1})=\zeta(s')\upharpoonright (\{\alpha\}\times k_{n-1}) \end{align*} $$

for some $s'\in \mathcal {S}$ such that $s'\subsetneq s$ . Therefore, the desired conclusion follows from the inductive hypothesis. The case $k\in [k_{n-1},k_n)$ is impossible, since $\phi (n-1)=s(n-1)=0$ implies that $[k_{n-1},k_n)\cap z_\alpha =\varnothing $ by our choice of $w_\alpha $ . To conclude the proof, assume that $k\in [k_n,\infty )$ . By picking $q\in G$ as above and using the assumption that $n>n^{p_\alpha }$ , one sees that $v_\alpha \setminus k_n\subseteq \pi _\alpha \big (\zeta (s)\big )$ by condition $(13)$ . Since $z_\alpha \subseteq v_\alpha $ , it follows that $k\in \pi _\alpha \big (\zeta (s)\big )$ .

Question 4.3. Let $\kappa <\mathfrak {p}$ be a cardinal, and let $X_\alpha $ be separable metrizable spaces with the $\mathsf {MP}$ for $\alpha \in \kappa $ . Does it follow that $\prod _{\alpha \in \kappa }X_\alpha $ has the $\mathsf {MP}$ ?

Proof. Set $X=\prod _{\alpha \in \kappa }\mathcal {F}_\alpha $ , and denote the projections by $\pi _\alpha :X\longrightarrow \mathcal {F}_\alpha $ for $\alpha \in \kappa $ . Assume that X is $\mathsf {CDH}$ . By [Reference Engelking5, Exercise 2.3.F.(c)], we must have $\kappa \leq \mathfrak {c}$ , otherwise X would not be separable. It now follows from [Reference Medini and Steprāns18, Theorem 8.1] that $\kappa <\mathfrak {p}$ .

It remains to show that each $\mathcal {F}_\alpha $ is a non-meager $\mathsf {P}$ -filter. By Theorem 1.1, it will suffice to show that each $\mathcal {F}_\alpha $ has the $\mathsf {MP}$ . In fact, since X contains closed copies of each $\mathcal {F}_\alpha $ , it will be enough to show that X has the $\mathsf {MP}$ . First assume that $\kappa $ is finite. In this case, since X is homeomorphic to a filter by Proposition 3.3, the desired conclusion follows from Theorem 1.1.

Now assume that $\kappa $ is infinite. In this case, it follows from Theorem 2.1 that X is a Baire space, hence non-meager. Since each $\pi _\alpha $ is open, it follows that each $\mathcal {F}_\alpha $ is non-meager. To see that X has the $\mathsf {MP}$ , pick a countable crowded $Q\subseteq X$ . Fix countable dense subsets D and E of X such that $D\supseteq Q$ and $E\subseteq \mathsf {Cof}^\kappa $ . Since X is $\mathsf {CDH}$ , there exists a homeomorphism $h:X\longrightarrow X$ such that $h[D]=E$ . Set $R=h[Q]\subseteq \mathsf {Cof}^\kappa $ . Since every subset of $\mathsf {Cof}$ has $\omega $ as a pseudointersection, we can apply Lemma 4.1, which yields a crowded $R'\subseteq R$ whose closure in X is compact. To conclude the proof, set $Q'=h^{-1}[R']$ , and observe that $Q'\subseteq Q$ . Since h is a homeomorphism, it is clear that $Q'$ is crowded and its closure in X is compact.

Proof. In order to prove the implication $(1)\rightarrow (2)$ , assume that condition $(1)$ holds. Fix a metrizable compactification $\gamma X$ of X; we will use $\mathsf {cl}$ to denote closure in $\gamma X$ . Pick a countable dense $D\subseteq X$ , and a closed copy N of $\omega ^\omega $ in $\gamma X\setminus D$ . Set $Q=\mathsf {cl}(N)\cap D$ , and observe that Q is crowded by Lemma 2.4. Therefore, by the $\mathsf {MP}$ , there exists a crowded $Q'\subseteq Q$ such that $\mathsf {cl}(Q')\subseteq X$ . Set $N'=\mathsf {cl}(Q')\setminus D\subseteq X\setminus D$ . Using Theorem 2.2, one sees that $N'\subseteq N$ is a closed copy of $\omega ^\omega $ in $\gamma X\setminus D$ , as desired.

The implication $(2)\rightarrow (3)$ is trivial. In order to prove the implication $(3)\rightarrow (1)$ , assume that condition $(3)$ holds. Pick a countable crowded $Q\subseteq X$ . Let $D\supseteq Q$ be a countable dense subset of X. Using [Reference Engelking5, Exercise 7.4.17], it is possible to find a zero-dimensional Polish subspace Z of $\gamma X$ such that $Z\cap D=Q$ . Since Polish spaces have the $\mathsf {MP}$ by [Reference Medini and Zdomskyy19, Proposition 2.3], there exists a compact subspace K of Z such that $K\cap Q$ is crowded and dense in K; we will use $\mathsf {cl}$ to denote closure in K. Set $N=K\setminus Q$ , and observe that N is homeomorphic to $\omega ^\omega $ by Theorem 2.2. Furthermore, the fact that $N=(\gamma X\setminus D)\cap K$ shows that N is closed in $\gamma X\setminus D$ . It follows from condition $(3)$ that there exists a closed copy $N'\subseteq N$ of $\omega ^\omega $ in $\gamma X\setminus D$ such that $N'\subseteq X\setminus D$ . Set $Q'=\mathsf {cl}(N')\cap Q$ , and observe that $Q'$ is crowded by Lemma 2.4. On the other hand $\mathsf {cl}(Q')\subseteq \mathsf {cl}(N')\subseteq N'\cup D\subseteq X$ , which shows that $Q'$ has compact closure in X, as desired.

Proof. Simply combine Theorem 1.1 and Proposition 6.1.

Proof. Using Theorem 2.2, one sees that $2^\omega \setminus D$ is homeomorphic to $\omega ^\omega $ for every countable dense subset D of $2^\omega $ . The rest of the proof is clear by Corollary 6.2.

Proof. Set $X=\bigcap _{\alpha \in \kappa }X_\alpha $ , and pick a countable crowded $Q\subseteq X$ . Fix a metrizable compactification $\gamma X$ of X. First, using [Reference Engelking5, Exercise 7.4.17], find a zero-dimensional Polish subspace W of $\gamma X$ such that $Q\subseteq W$ . Since Polish spaces have the $\mathsf {MP}$ by [Reference Medini and Zdomskyy19, Proposition 2.3], there exists a compact subspace K of W such that $K\cap Q$ is crowded and dense in K; we will use $\mathsf {cl}$ to denote closure in K.

Set $X'=X\cap K$ and $X^{\prime }_\alpha =X_\alpha \cap K$ for $\alpha \in \kappa $ . Observe that each $X^{\prime }_\alpha $ has the $\mathsf {MP}$ because it is a closed subspace of $X_\alpha $ . Furthermore, it is clear that K is a metrizable compactification of each $X^{\prime }_\alpha $ because $K\cap Q\subseteq X^{\prime }_\alpha $ . It follows from Proposition 6.1 that each $X^{\prime }_\alpha \setminus Q$ is Miller-full in $K\setminus Q$ . Since $K\setminus Q$ is homeomorphic to $\omega ^\omega $ by Theorem 2.2, the assumption $\kappa <\mathsf {add}(m^0)$ ensures that $X'\setminus Q$ is Miller-full in $K\setminus Q$ . In particular, there exists a closed copy N of $\omega ^\omega $ in $K\setminus Q$ such that $N\subseteq X'\setminus Q$ . Set $Q'=\mathsf {cl}(N)\cap Q$ , and observe that $Q'$ is crowded by Lemma 2.4. On the other hand $\mathsf {cl}(Q')\subseteq \mathsf {cl}(N)\subseteq N\cup Q\subseteq X$ , which shows that $Q'$ has compact closure in X, as desired.

Question 7.2. Set $\kappa =\mathsf {add}(m^0)$ . Are there a separable metrizable space Z and subspaces $X_\alpha $ of Z with the $\mathsf {MP}$ for $\alpha \in \kappa $ such that $\bigcap _{\alpha \in \kappa }X_\alpha $ does not have the $\mathsf {MP}$ ?

Proof. Fix a metrizable compactification $\gamma X$ of X. We will show that condition  $(3)$ of Proposition 6.1 holds for the compactification $\gamma X\times K$ of $X\times K$ ; we will use $\mathsf {cl}$ to denote closure in $\gamma X\times K$ . So pick a countable dense subset D of $X\times K$ , and a closed copy N of $\omega ^\omega $ in $(\gamma X\times K)\setminus D$ . Denote by $\pi :\gamma X\times K\longrightarrow \gamma X$ the projection, and set $E=\pi [D]$ . Given $x\in E$ , set $D_x=(\mathsf {cl}(N)\setminus N)\cap (\{x\}\times K)$ , and observe that each $D_x\subseteq D$ .

Case 1. There exists $x\in E$ such that $D_x$ is non-scattered.

Let $Q\subseteq D_x$ be crowded. Set $N'=\mathsf {cl}(Q)\setminus D$ , and observe that $N'\subseteq N$ because $\mathsf {cl}(Q)\subseteq \mathsf {cl}(N)\subseteq N\cup D$ . Furthermore, it is clear that $N'$ is closed in $(\gamma X\times K)\setminus D$ . At this point, one sees that $N'$ is homeomorphic to $\omega ^\omega $ by Theorem 2.2. Finally, the fact that $\mathsf {cl}(Q)\subseteq \{x\}\times K\subseteq X\times K$ shows that $N'\subseteq (X\times K)\setminus D$ .

Case 2. $D_x$ is scattered for every $x\in E$ .

Since scattered separable metrizable spaces are Polish (see [Reference Kechris13, Corollary 21.21]), we see that $(\{x\}\times K)\setminus D_x$ must be $\sigma $ -compact for every $x\in E$ . Hence,

$$ \begin{align*}N\cap (\{x\}\times K)=\big((\{x\}\times K)\setminus D_x\big)\cap\mathsf{cl}(N) \end{align*} $$

is $\sigma $ -compact for every $x\in E$ .

Set $Z=(\gamma X\setminus E)\times K$ , and observe that $N\setminus (E\times K)$ is an analytic subset of the Polish space Z. Assume, in order to get a contradiction, that $N\setminus (E\times K)$ is contained in a $\sigma $ -compact subspace of Z. Since E is countable and $N\cap (\{x\}\times K)$ is $\sigma $ -compact for every $x\in E$ , it follows that N itself is $\sigma $ -compact, a contradiction. Therefore, by Theorem 2.3, we can assume without loss of generality that N is a (closed) subset of Z.

Now set $W=\gamma X\setminus E$ , and observe that $\pi [N]$ is an analytic subset of the Polish space W. If $\pi [N]$ were contained in a $\sigma $ -compact subspace of W, then it would easily follow that N is $\sigma $ -compact, a contradiction. Therefore, by Theorem 2.3, there exists a closed copy M of $\omega ^\omega $ in W such that $M\subseteq \pi [N]$ . But X has the $\mathsf {MP}$ , so by Proposition 6.1 there exists a closed copy $M'\subseteq M$ of $\omega ^\omega $ in W such that $M'\subseteq X\setminus E$ .

Finally, set $N'=N\cap \pi ^{-1}[M']$ , and observe that $N'$ is a closed subset of Z. Therefore, if $N'$ were contained in a $\sigma $ -compact subspace of Z, then $N'$ itself would be $\sigma $ -compact, contradicting the fact that $\pi [N']=M'$ . It follows from Theorem 2.3 that there exists a closed copy $N"$ of $\omega ^\omega $ in Z such that $N"\subseteq N'$ . It is easy to realize that $N"$ is a closed copy of $\omega ^\omega $ in $(\gamma X\times K)\setminus D$ such that $N"\subseteq (X\times K)\setminus D$ .

Proof. Fix metrizable compactifications $\gamma X_\alpha $ of $X_\alpha $ for $\alpha \in \kappa $ . Set $Z=\prod _{\alpha \in \kappa }\gamma X_\alpha $ , and observe that Z is separable metrizable because $\kappa \leq \omega $ . Define

$$ \begin{align*}X^{\prime}_\alpha=\{x\in Z:x(\alpha)\in X_\alpha\} \end{align*} $$

for $\alpha \in \kappa $ , and observe that each $X^{\prime }_\alpha $ has the $\mathsf {MP}$ by Lemma 8.1. Since clearly $\prod _{\alpha \in \kappa }X_\alpha =\bigcap _{\alpha \in \kappa }X^{\prime }_\alpha $ , the desired conclusion follows from Theorem 7.1.