Fact 2.5 If ${\mathcal {F}}$ is a front on a countable set M, there exists $N\in [M]^\omega $ such that ${\mathcal {F}}|N$ is a barrier.

Fact 2.6 For every barrier ${\mathcal {B}}$ on M, if we partition ${\mathcal {B}}={\mathcal {B}}_0\sqcup \cdots \sqcup {\mathcal {B}}_n$ , there are $N\in [M]^\omega $ and $i\leq n$ such that ${\mathcal {B}}_i|N$ is a barrier.

Proof Let $s\in [\omega ]^{<\omega }$ and $b\in {\mathcal {B}}$ such that $s\subseteq b$ . Define $M=s\cup (\omega \setminus \max (b)+1)$ . We can find $b'\in {\mathcal {B}}$ such that $b'\sqsubseteq M$ , and hence, $s\sqsubseteq b'$ since otherwise, $b'\sqsubset s\subseteq b$ would contradict that ${\mathcal {B}}$ is a $\subseteq $ -antichain.

Proof Fix $a=\{a_0,\ldots ,a_{n-1}\}$ and $b=\{b_0,\ldots ,b_{k-1}\}$ as in the lemma with $k<n$ . We will prove the Lemma by induction on k.

If $k=1$ , then $b=\{b_0\}$ for some $b_0\geq a_{n-1}$ . Let $M\in [\omega ]^\omega $ with $a\ast b\sqsubseteq M$ . There is an $s\in {\mathcal {B}}$ such that $s\sqsubseteq M$ . Notice that $|s|\geq n$ since otherwise, $s\sqsubset a$ would contradict that ${\mathcal {B}}$ is Sperner. Thus, $a\ast b\sqsubseteq s\in {\mathcal {B}}$ .

Now let $k>1$ and assume the result is true for ${k-1}$ . Pick any $M\in [\omega ]^\omega $ such that $a\ast b\sqsubseteq M$ and let $s\in {\mathcal {B}}$ be an initial segment of M. If $|s|<n$ , then $s\sqsubseteq \{a_0,\ldots ,a_{n-k-1},b_0,\ldots ,b_{k-2}\}\subseteq \{a_0,\ldots ,a_{n-k},b_0,\ldots ,b_{k-2}\}=a\ast b'$ , where $b'=\{b_0,\ldots ,b_{k-2}\}$ . Since $|b'|=k-1$ , by the inductive hypothesis, we get that $s\subsetneq a\ast b'\sqsubseteq s'$ for some $s'\in {\mathcal {B}}$ , contradicting that ${\mathcal {B}}$ is Sperner. Therefore, $|s|\geq n$ , and hence, $a\ast b\sqsubseteq s\in {\mathcal {B}}$ .

Proof Let ${\mathcal {B}}$ be a barrier of rank k. If there is $a\in {\mathcal {B}}$ of size l, then $\rho ({\mathcal {B}})\geq l$ . Thus, ${\mathcal {B}}\subseteq [\omega ]^{\leq k}$ . It is also easy to see that ${\mathcal {B}}\cap [\omega ]^k\neq \emptyset $ since otherwise, $\rho ({\mathcal {B}})<k$ .

Take any $a\in {\mathcal {B}}\cap [\omega ]^k$ and let $m=\max (a)+1$ . By Lemma 2.9, if $b\in {\mathcal {B}}|(\omega \setminus m)$ has size less than k, we can find $s\in {\mathcal {B}}$ such that $b'=a\ast b\sqsubseteq s$ . As $|b|<k$ , we have that $b\subsetneq b'\sqsubseteq s$ , which contradicts that ${\mathcal {B}}$ is Sperner. Hence, ${\mathcal {B}}|(\omega \setminus m)\subseteq [\omega \setminus m]^k$ .

However, for every $b\in [\omega \setminus m]^k$ , we can take an infinite set M such that $b\sqsubset M$ . Thus, there is an initial segment of M in ${\mathcal {B}}$ , and from the previous inclusion, it must be the case that $b\in {\mathcal {B}}$ . Therefore, ${\mathcal {B}}|(\omega \setminus m)=[\omega \setminus m]^k$ .

Notation 2.11 Given a family ${\mathcal {B}}\subseteq [\omega ]^{<\omega }$ , $s\in [\omega ]^{<\omega }$ , and $n\in \omega $ , set

• • ${\mathcal {B}}(s)=\{t\in {\mathcal {B}}:s\sqsubseteq t\}$ ,

• • ${\mathcal {B}}[s]=\{t\setminus s:t\in {\mathcal {B}}(s)\}$ ,

• • $s^{\smallfrown {\mathcal {B}}}=\{s\cup t:t\in {\mathcal {B}}\ \land \ \min (t)>\max (s)\}$ ,

• • ${\mathcal {B}}+n=\{s+n:s\in {\mathcal {B}}\}$ , where $s+n=\{m+n:m\in s\}$ .

Proof By definition, we can find $m_i$ such that $\rho (\{m_i\})\geq i$ for every $i\in \omega $ . By Corollary 2.10, we can also find $k_i$ for every $i\in \omega $ , such that ${\mathcal {B}}[{m_i}]=[\omega \setminus k_i]^{n_i}$ , where $n_i=\rho (\{m_i\})$ . Notice that we can pick $\{m_i:i\in \omega \}$ increasing and such that $m_{i+1}>k_i$ . It is now easy to see that $M=\{m_i:i\in \omega \}$ works.

Notation 2.17 Given two families ${\mathcal {B}},{\mathcal {C}}\subseteq [\omega ]^{<\omega }$ , we denote by ${\mathcal {B}}\sqsubseteq {\mathcal {C}}$ if

• • $\forall s\in {\mathcal {B}}\ \exists t\in {\mathcal {C}}\ (s\sqsubseteq t)$ ,

• • $\forall t\in {\mathcal {C}}\ \exists s\in {\mathcal {B}}\ (s\sqsubseteq t).$

Proof Define ${\mathcal {B}}_0=\{b\in {\mathcal {B}}:\exists c\in {\mathcal {C}}(b\subseteq c)\}$ . Since ${\mathcal {B}}={\mathcal {B}}_0\cup ({\mathcal {B}}\setminus {\mathcal {B}}_0)$ , we can find, by Nash-Williams theorem, an infinite set $M\in [N]^\omega $ such that either $({\mathcal {B}}\setminus {\mathcal {B}}_0)|M=\emptyset $ or ${\mathcal {B}}_0|M=\emptyset $ . If $({\mathcal {B}}\setminus {\mathcal {B}}_0)|M=\emptyset $ , then we are done, as this implies that ${\mathcal {B}}|M\sqsubseteq {\mathcal {C}}|M$ . To see this, notice that for every $b\in {\mathcal {B}}|M$ ( and hence $b\in {\mathcal {B}}_0$ ), there is $c\in {\mathcal {C}}$ such that $b\subseteq c$ , but if we define $X=b\cup (M\setminus \max (b))$ , there is also $c'\in {\mathcal {C}}$ such that $c'\sqsubseteq X$ . As $b\subseteq c\in {\mathcal {C}}$ , it happens that $b\sqsubseteq c'$ and $c'\in {\mathcal {C}}|M$ . For $c\in {\mathcal {C}}|M$ , we can also find $b\in {\mathcal {B}}|M$ such that $b\sqsubseteq c\cup (M\setminus (\max (c)+1))$ . The case where $c\sqsubset b$ is not possible since $b\in {\mathcal {B}}_0$ , and thus, there is another $c'\in {\mathcal {C}}|M$ such that $b\subseteq c'$ .

Assume otherwise ${\mathcal {B}}_0|M=\emptyset $ and define ${\mathcal {C}}_0=\{c\in {\mathcal {C}}|M:\exists b\in {\mathcal {B}}|M(c\subseteq b)\}$ . Again, by Nash-Williams theorem, we can find $M'\in [M]^\omega $ such that either ${\mathcal {C}}_0|M'=\emptyset $ or $({\mathcal {C}}\setminus {\mathcal {C}}_0)|M'=\emptyset $ . The case ${\mathcal {C}}_0|M'=\emptyset $ is impossible since we can find $b\in {\mathcal {B}}|M'\subseteq {\mathcal {B}}|M$ and $c\in {\mathcal {C}}|M'$ such that $b,c\sqsubseteq M'$ and then either $b\subseteq c$ or $c\subseteq b$ contradicting the choices of M and $M'$ . Thus, $({\mathcal {C}}\setminus {\mathcal {C}}_0)|M'=\emptyset $ , and as in the previous case, we get that ${\mathcal {C}}|M'\sqsubseteq {\mathcal {B}}|M'$ .

For the last assertion of the statement, notice that if ${\mathcal {C}}|M\sqsubseteq {\mathcal {B}}|M$ , then $T({\mathcal {C}}|M)\subseteq T({\mathcal {B}}|M)$ . In particular,

$$ \begin{align*}\rho({\mathcal{C}})=\rho({\mathcal{C}}|M)=\rho_{T({\mathcal{C}}|M)}(\emptyset)\leq\rho_{T({\mathcal{B}}|M)}(\emptyset)=\rho({\mathcal{B}}|M)\leq\rho({\mathcal{B}}),\end{align*} $$

which finishes the proof.

Proof Let us alternately define two sequences $\{a_i:i\in \omega \}$ and $\{b_i:i\in \omega \}$ such that $a_i<b_i<a_{i+1}$ for every $i\in \omega $ . We will also denote by $\{k_i:i\in \omega \}$ the increasing sequence defined by these two sequences; that is, $k_{2i}=a_i$ and $k_{2i+1}=b_i$ . We will also simultaneously define $f\in \omega ^\omega $ which is completely determined by the sequence $\{k_i:i\in \omega \}$ as follows: $f\upharpoonright [0,k_0)=0$ and $f(n)=i$ if and only if $n\in [k_i,k_{i+1})$ . We shall show that f satisfies the definition of ${\mathcal {C}}\preceq {\mathcal {B}}$ . To save notation, let us write $\rho _{\mathcal {B}}$ and $\rho _{\mathcal {C}}$ instead of $\rho _{T({\mathcal {B}})}$ and $\rho _{T({\mathcal {C}})}$ , respectively. Also, given $s\in [\omega ]^\omega \setminus T({\mathcal {C}})$ , we will define $\rho _{\mathcal {C}}(s)=-1$ so that we can talk about the rank of s even if s is not in $T({\mathcal {C}})$ .

Step 0:

Since $\rho _{\mathcal {B}}(\emptyset )\geq \rho _{\mathcal {C}}(\emptyset )$ , there are $a_0<b_0<\omega $ such that $\rho _{\mathcal {B}}(\{a_0\})\geq \rho _{\mathcal {C}}(\{0\})$ and $\rho _{\mathcal {B}}(\{b_0\})\geq \rho _{\mathcal {C}}(\{1\})$ .

Now assume we have defined $\{a_i:i\leq n\}$ and $\{b_i:i\leq n\}$ .

Step a:

Inductive hypothesis at n:

Let us call

$$ \begin{align*}S_a(n)=\Big\{s\in T({\mathcal{B}}):s\subseteq\bigcup_{i< n}[a_i,b_i)\ \land\ \forall i< n(|s\cap [a_i,b_i)|\leq 1)\Big\}.\end{align*} $$

We will ensure that the following inductive hypothesis will hold throughout the construction: Given $s\in S_a(n)$ ,

1. ϕ _a (n, s) $\equiv \rho _{\mathcal {B}}(s)\geq \rho _{\mathcal {C}}(f[s])$

2. ψ _a (n, s) $\equiv $ if $j\geq a_n$ , then $\rho _{\mathcal {B}}(s\cup \{j\})\geq \rho _{\mathcal {C}}(f[s\cup \{2n\}])$ .

We will also write $\phi _a(n)$ as a shorthand for $\forall s\in S_a(n)\phi _a(n,s)$ and $\psi _a(n)$ as a shorthand for $\forall s\in S_a(n)\psi _a(n,s)$ . It is worth pointing out that $S_a(n)$ , $\phi _a(n,s)$ and $\psi _a(n,s)$ do not mention $b_n$ . Note that if $j\geq a_0$ , we have that $\rho _{\mathcal {B}}(\{j\})\geq \rho _{\mathcal {B}}(\{a_0\})\geq \rho _{\mathcal {C}}(\{0\})$ as ${\mathcal {B}}$ is uniform. Thus, $\psi _a(0)$ is satisfied, and $\phi _a(0)$ is vacuously satisfied too (here, $S_a(0)=\{\emptyset \}$ ).

Construction of a _n+1:

We need to take care of all $s\in S_a(n+1)$ (note that this set is well-defined since it only depends on $\{a_i,b_i:i\leq n\}$ ).

First, consider $s\in S_a(n)$ . By $\phi _a(n,s)$ , we have that $\rho _{\mathcal {B}}(s)\geq \rho _{\mathcal {C}}(f[s])$ , and thus, we can find $m_s^a$ such that

1. (⋆₁ ^a ) $\rho _{\mathcal {B}}(s\cup \{m\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+2\})$ for every $m\geq m_s^a$ .

Otherwise, $s\in S_a(n+1)\setminus S_a(n)$ , and we can write $s'=s\setminus \{j\}$ , where $j=\max (s)\geq a_n$ and $s'\in S_a(n)$ . Then, we know that $\rho _{\mathcal {B}}(s'\cup \{j\})\geq \rho _{\mathcal {C}}(f[s']\cup \{2n\})$ by $\psi _a(n,s)$ . Since $j\in [a_n,b_n)$ , we have that $f(j)=2n$ . Thus,

1. (*^a) $\rho _{\mathcal {B}}(s)=\rho _{\mathcal {B}}(s'\cup \{ j\})\geq \rho _{\mathcal {C}}(f[s']\cup \{f(j)\})=\rho _{\mathcal {C}}(f[s'\cup \{j\}])=\rho _{\mathcal {C}}(f[s]).$

Then, there is $m_s^a\in \omega $ such that $\rho _{\mathcal {B}}(s\cup \{m_s^a\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+2\})$ . Since ${\mathcal {B}}$ is uniform,

1. (⋆₂ ^a ) $\rho _{\mathcal {B}}(s\cup \{m\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+2\})$ for every $m\geq m_s^a$ .

Finally, define

$$ \begin{align*}a_{n+1}=\max\{m_s^a\mid s\in S_a(n+1) \}.\end{align*} $$

Inductive hypothesis for n + 1:

We want to see that $\phi _a(n+1)$ and $\psi _a(n+1)$ hold. Let $s\in S_a(n+1)$ . If ${s\in S_a(n)}$ , then $\phi _a(n+1,s)$ follows from $\phi _a(n,s)$ . In consequence, we can assume that ${s\in S_a(n+1)\setminus S_a(n)}$ and let $s'=s\setminus \{j\}$ , where $j=\max (s)$ . Hence, by $(\ast ^a)$ , we have that $\rho _{{\mathcal {B}}}(s)\geq \rho _{{\mathcal {C}}}(f[s])$ , which means that $\phi _n(n+1,s)$ holds and, consequently, $\phi _a(n+1)$ does.

Now let us see that $\psi _a(n+1)$ is also true. Let $s\in S_a(n+1)$ and $j\geq a_{n+1}$ . If ${s\in S_a(n)}$ , then $m_a^s$ satisfies $(\star _1^a)$ , and if $s\in S_a(n+1)\setminus S_a(n)$ , we have that $m_a^s$ satisfies $(\star _2^a)$ . In either case, as $a_n+1\geq m_a^s$ , we have that $\psi _a(n+1,s)$ holds, and so does $\psi _a(n+1)$ .

Step b:

Inductive hypothesis at n:

The construction is dual to that of part a. Let us start by calling

$$ \begin{align*}S_b(n)=\Big\{s\in T({\mathcal{B}}):s\subseteq\bigcup_{i< n}[b_i,a_{i+1})\ \land\ \forall i<n(|s\cap [b_i,a_{i+1})|\leq 1)\Big\}.\end{align*} $$

The corresponding inductive formulas for $s\in S_b(n)$ are now

1. ϕ _b (n, s) $\equiv \rho _{\mathcal {B}}(s)\geq \rho _{\mathcal {C}}(f[s])$

2. ψ _b (n, s) $\equiv $ if $j\geq b_n$ , then $\rho _{\mathcal {B}}(s\cup \{j\})\geq \rho _{\mathcal {C}}(f[s\cup \{2n+1\}])$ .

We will write again $\phi _b(n)$ and $\psi _b(n)$ as shorthands for $\forall s\in S_b(n)\phi _a(n,s)$ and $\forall s\in S_b(n)\psi _a(n,s)$ , respectively. In the particular case of $\phi _b$ , we have that it is the same formula as $\phi _a$ with different domain. Since $a_{n+1}$ has already been defined, the interval $[b_n,a_{n+1})$ considered in $S_b(n+1)$ makes complete sense.

Construction of b _n+1:

Given $s\in S_b(n)$ , it follows from $\phi _b(n,s)$ that $\rho _{\mathcal {B}}(s)\geq \rho _{\mathcal {C}}(f[s])$ , and thus, we can find $m_s^b$ such that

1. (⋆₁ ^b ) $\rho _{\mathcal {B}}(s\cup \{m\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+3\})$ for every $m\geq m_s^b$ .

Otherwise, $s\in S_b(n+1)\setminus S_b(n)$ , and we can write $s'=s\setminus \{j\}$ , where $j=\max (s)\geq b_n$ and $s'\in S_b(n)$ . Thus, $\rho _{\mathcal {B}}(s'\cup \{j\})\geq \rho _{\mathcal {C}}(f[s']\cup \{2n+1\})$ by $\psi _b(n,s)$ . Since $j\in [b_n,a_{n+1})$ , we have that $f(j)=2n+1$ . Then,

1. (* ^b ) $\rho _{\mathcal {B}}(s)=\rho _{\mathcal {B}}(s'\cup \{ j\})\geq \rho _{\mathcal {C}}(f[s']\cup \{f(j)\})=\rho _{\mathcal {C}}(f[s'\cup \{j\}])=\rho _{\mathcal {C}}(f[s]).$

We can again find $m_s^b\in \omega $ such that $\rho _{\mathcal {B}}(s\cup \{m_s^b\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+3\})$ , and moreover,

1. (⋆₂ ^b ) $\rho _{\mathcal {B}}(s\cup \{m\})\geq \rho _{\mathcal {C}}(f[s]\cup \{2n+3\})$ for every $m\geq m_s^b$ .

We then define

$$ \begin{align*}b_{n+1}=\max\{m_s^b\mid s\in S_b(n+1) \}.\end{align*} $$

Inductive hypothesis for n + 1:

If $s\in S_b(n)$ , then $\phi _b(n+1,s)$ follows from $\phi _b(n,s)$ , and otherwise, $s\in S_b(n+1)\setminus S_b(n)$ can be written as $s'=s\setminus \{j\}$ , where $j=\max (s)$ . By $(\ast ^b)$ , we have that $\rho _{{\mathcal {B}}}(s)\geq \rho _{{\mathcal {C}}}(f[s])$ , and then $\phi _b(n+1)$ holds.

To see that $\psi _b(n+1)$ is true, let $s\in S_b(n+1)$ and $j\geq b_{n+1}$ . Then, $b_{n+1}\geq m_s^b$ , and either by $(\star _1^b)$ or $(\star _2^b)$ , we have that $\phi _b(n+1,s)$ holds.

This finishes the construction, and it remains to prove that f works.

Clearly, f is finite-to-one and nondecreasing. Given $M\in [\omega ]^\omega $ , we can find an infinite subset $N\in [M]^\omega $ such that $|N\cap [k_i,k_{i+1}]|\leq 1$ for every $i\in \omega $ , and either $N\cap \bigcup _{i\in \omega }[k_{2i},k_{2i+1})=\emptyset $ or $N\cap \bigcup _{i\in \omega }[k_{2i+1},k_{2i+2})=\emptyset $ . Without loss of generality, assume that $N\subseteq \bigcup _{i\in \omega }[k_{2i},k_{2i+1})=\bigcup _{i\in \omega }[a_i,b_i)$ .

Let now $b\in {\mathcal {B}}|N$ . It follows from the choice of N that there is an $n\in \omega $ such that $b\in S_a(n)$ . By $\phi _a(n,b)$ , we know that $0=\rho _{\mathcal {B}}(b)\geq \rho _{\mathcal {C}}(f[b])$ . Necessarily, $\rho _{\mathcal {C}}(f[b])\in \{0,-1\}$ , but in either case, we can conclude that there is $c\in {\mathcal {C}}$ such that $c\sqsubseteq f[b]$ . This finishes the proof.

Proof By the fact that the preorder $\preceq $ is transitive and the previous proposition, it suffices to show that ${\mathcal {S}}\preceq {\mathcal {B}}$ , where ${\mathcal {S}}$ is the Schreier barrier.

Since ${\mathcal {B}}$ has rank $\omega $ , it follows from Lemma 2.9 that for every $i\in \omega $ , we can find $m_i$ such that $\rho _{\mathcal {B}}(m)\geq i+1$ for every $m\geq m_i$ . Let $f\in \omega ^\omega $ be given by $f(n)=i$ if and only if $n\in [m_i,m_{i+1})$ (and we define $f\upharpoonright m_0$ arbitrarily). We claim that f witnesses that ${\mathcal {S}}\preceq {\mathcal {B}}$ .

Indeed, given $M\in [\omega ]^\omega $ , we can find $N_0\in [M]^\omega $ such that $N_0\cap [0,m_0)=\emptyset $ and $|N_0\cap [m_i,m_{i+1})|\leq 1$ for every $i\in \omega $ . Let $n_0=\min (N_0)$ , and for every $i\in \omega $ , define $n_{i+1}$ as follows: for each $j<i+1$ , we can apply Corollary 2.10 and find $k_j$ such that ${\mathcal {B}}[n_j]=[\omega \setminus k_j]^r$ for some $r\geq l$ , where l is the unique natural number such that ${n_j\in [m_l,m_{l+1})}$ . Pick $n_{i+1}>k_j$ for every $j<i+1$ . Thus, $N=\{n_i:i\in \omega \}\in [M]^\omega $ .

Fix $b\in ({\mathcal {B}}|N)$ and let $n_i=\min (b)$ . It follows from the definitions that $b\setminus \{n_i\}\in [\omega \setminus k_j]^r$ for some $r\geq l$ , where $n_i\in [m_l,m_{l+1})$ . Then, $|f[b]|\geq l+1$ as f is one to one on N, but $f(n_i)=l$ and ${\mathcal {S}}(l)=\{s\in [\omega ]^{<\omega }:\min (s)=l\land |s|=l+1\}$ . Thus, we can conclude that $c\sqsubseteq f[b]$ defined as the first $l+1$ elements of $f[b]$ is an element of ${\mathcal {S}}$ .

Proof Assume X is ${\mathcal {B}}$ sequentially compact and let $f\in \omega ^\omega $ as in the definition of ${\mathcal {C}}\preceq {\mathcal {B}}$ . Let $h:{\mathcal {C}}\to X$ and define $\hat {h}:{\mathcal {B}}\to X$ by $\hat {h}(b)=h(c)$ for any $c\in {\mathcal {C}}$ that is $\sqsubseteq $ -compatible with $f[b]$ . Notice that since ${\mathcal {C}}$ is a barrier, there is at least one of such c, and then $\hat {h}$ is well-defined.

Since X is ${\mathcal {B}}$ -sequentially compact, there is $M\in [\omega ]^\omega $ such that $\hat {h}\upharpoonright ({\mathcal {B}}|M)$ converges to some $x\in X$ . Take $N\in [M]^\omega $ such that

$$ \begin{align*}\forall b\in({\mathcal{B}}|N)\ \exists c\in{\mathcal{C}}\ (c\sqsubseteq f[b])\end{align*} $$

given by the definition of ${\mathcal {C}}\preceq {\mathcal {B}}$ , and define $M_0=f[N]$ . It turns out that $\hat {h}\upharpoonright ({\mathcal {B}}|N)$ also converges to x. We shall prove that $h\upharpoonright ({\mathcal {C}}|M_0)$ converges to the same $x\in X$ .

To see this, fix an open neighborhood $U\in {\mathcal {N}}(x)$ . We can find $n\in \omega $ such that $\hat {h}(b)\in U$ whenever $b\in {\mathcal {B}}|(N\setminus n)$ . Take $c\in {\mathcal {C}}|(M_0\setminus f(n))$ . Thus, $c=\{f(m_0),\ldots ,f(m_j)\}$ for some set $a=\{m_i:i\leq j\}\subseteq N\setminus n$ . Take any infinite set $N'\subseteq N$ such that $a\sqsubseteq N'$ . We can find $b\in {\mathcal {B}}|N$ such that $b\sqsubseteq N'$ . It follows that $a\sqsubseteq b$ since otherwise, $f[b]\sqsubset f[a]=c$ would contradict that $b\subseteq N$ . Therefore, $h(c)=\hat {h}(b)\in U$ , as desired.

Proof That (3) implies (1) follows directly from Proposition 2.20 and Theorem 3.4. The other implications are clear from the definitions.

Proof By Corollary 3.5, it is enough to prove that X is $({\mathcal {B}}|M)$ -sequentially compact. Let $F:=\omega \setminus M$ . Regardless of whether F is finite or X is not $({\mathcal {B}}|F)$ -sequentially compact, there exists a function $g:{\mathcal {B}}|F\to X$ without infinite convergence subsequences (here, convergence means convergence with respect to barriers). Now let $f:{\mathcal {B}}|M\to X$ . We want to prove that f admits an infinite convergent subsequence $f\upharpoonright ({\mathcal {B}}|M')$ for some $M'\in [\omega ]^\omega $ . For this, let $\hat {f}:{\mathcal {B}}\to X$ be any common extension of both f and g. As X is ${\mathcal {B}}$ -sequentially compact, there is $N\in [\omega ]^\omega $ such that $\hat {f}\restriction N$ is convergent. Now $N\cap F$ should be finite, since otherwise, $\hat {f}\restriction (N\cap F)$ would also be an infinite convergent subsequence for g. This way, $N':=N\setminus F$ is infinite, and thus, ${\hat {f}\restriction N'=f\restriction N'}$ is an infinite convergent subsequence for f.

Proof We will show by induction that X is $\alpha $ -sequentially compact for every ${\alpha <\omega _1}$ . If $\alpha =1$ , then X being sequentially compact is equivalent to X being 1-sequentially compact, as the only uniform barrier of rank $1$ is $[\omega ]^1$ .

We proceed to prove the inductive step at $\beta $ , regardless of it being a successor ordinal, and hence of the form $\beta =\alpha +1$ , or a limit ordinal, which would then be the $\operatorname {sup}\alpha _n$ , where $\alpha _n$ form an increasing sequence. Now let $f:{\mathcal {B}}\to X$ be any function where ${\mathcal {B}}$ is uniform of rank $\beta $ . For all $n\in \omega $ , ${\mathcal {B}}[n]$ is either an $\alpha $ -uniform barrier on $\omega \setminus (n+1)$ or an $\alpha _n$ -uniform barrier if $\beta $ were a limit ordinal.

For all $n\in \omega $ , let $f_n:{\mathcal {B}}[n]\to X$ be the function mapping $s\mapsto f(\{n\}\cup s)$ . By our inductive hypothesis, there is an infinite subset $N_0$ and a point $x_0\in X$ such that ${f_0\restriction ({\mathcal {B}}[0]|N_0)\rightarrow x_0}$ . Let $n_0=0$ and $n_1=\min N_0\setminus \{0\}$ .

Note that for any infinite subset A of $\omega $ and any $n\in \omega $ , ${\mathcal {B}}[n]|A$ is also uniform of rank $\alpha $ if $\beta =\alpha +1$ or of rank $\alpha _n$ if $\beta $ were a limit ordinal. Now suppose that we have defined a decreasing sequence of infinite subsets $N_0\supseteq N_1\supseteq \dots \supseteq N_{i-1}$ and an increasing sequence $n_0<n_1<\dots <n_i$ in addition to a subset $\{x_0,\dots ,x_{i-1}\}$ of X. Consider the function $f_{n_{i}}:({\mathcal {B}}[n_i]| N_{i-1})\rightarrow X$ . Applying our inductive hypothesis yields an infinite subset $N_{i}\subseteq N_{i-1}$ such that $f_{n_i}\restriction ({\mathcal {B}}[n_i]|N_{i})\rightarrow x_i$ , for some point $x_i\in X$ . We also define $n_{i+1}=\min (N_i\setminus \{n_0,\ldots ,n_i\})$ . Thereby, we obtain a decreasing sequence of infinite subsets $N_0\supseteq N_1\supseteq N_2\supseteq \dots $ , a subset $\{x_i:i\in \omega \}$ of the space X, and a strictly increasing sequence $\langle n_i:i\in \omega \rangle $ such that $f_{n_i}\restriction ({\mathcal {B}}[n_i]|N_i)\rightarrow x_i$ , where $n_{i+1}=\min N_i\setminus \{n_0,\ldots ,n_i\}$ for all $i\in \omega $ .

Since $\{x_i:i\in \omega \}\subseteq X$ and X is sequentially compact, there is a convergent subsequence, $Y=\{x_{i_j}:j\in \omega \}$ , with limit point $x.$ Re-numerating the indices, we can view $Y=\{x_i:i\in \omega \}$ as the convergent subsequence. Also, let $N=\{n_i:i\in \omega \}$ .

Take any open set U in X containing the point x. As $x_i\to x$ , there is an integer ${m_U\in \omega }$ such that for all $i>m_U$ , we have that $x_i\in U.$ Also, for all such i, there is another integer $\phi _U(n_i)$ such that

(*) $$ \begin{align} f_{n_i}[{\mathcal{B}}[n_i]| (N_i\setminus\phi_U(n_i))]\subseteq U. \end{align} $$

For any $j\leq m_U$ and any $j\notin \{n_i:i\in \omega \}$ , let $\phi _U(j)=0$ . Thus, any open subset U of X containing x induces a function $\phi _U:\omega \rightarrow \omega $ .

Let $\eta (x)$ be a local neighborhood base of x of minimum size. Since $\chi (X)<\mathfrak b$ , the set of functions $\{\phi _U:U\in \eta (x)\}$ is bounded by some increasing function $\psi :\omega \rightarrow \omega $ ; that is, $\phi _U<^*\psi $ for every $U\in \eta (x)$ .

We can find an increasing subsequence $\langle m_i:i\in \omega \rangle $ of $\langle n_i:i\in \omega \rangle $ such that $m_{i+1}>\psi (m_i)$ for every $i\in \omega $ . We claim that $f\upharpoonright ({\mathcal {B}}|M)\to x$ , where $M=\{m_i:i\in \omega \}$ .

Fix an open set $U\in \eta (x)$ and let $k\in \omega $ such that $x_j\in U$ and $\psi (j)>\phi _U(j)$ for every $j\geq k$ . Let $s\in {\mathcal {B}}|(M\setminus \psi (k))$ and let us write $s=\{m_{k_0},\ldots ,m_{k_n}\}$ and $s'=s\setminus \{m_{k_0}\}$ . Note that for every $i>0$ , we have that

$$ \begin{align*}m_{k_i}>\psi(m_{k_0})>\phi(m_{k_0}),\end{align*} $$

as $m_{k_0}\geq \psi (k)\geq k$ . In particular, $s'\cap \phi (m_{k_0})=\emptyset $ .

Now let $j\in \omega $ such that $m_{k_0}=n_j$ . Note that if $j'>j$ , then $n_{j'}\in N_j$ . As any $m_{k_i}>m_{k_0}$ for $i>0$ and this implies that $m_{k_i}=n_{j'}$ for some $j'>j$ , we can conclude that $s'\subseteq N_j$ .

Combining the previous arguments, we get that $s'\subseteq N_j\setminus \phi _U(n_j)$ , and we already know that $s=\{n_j\}\cup s'\in {\mathcal {B}}$ , which implies that $s'\in {\mathcal {B}}[n_j]$ . Therefore, ${f(s)=f_{n_j}(s')\in U}$ by ( $\ast $ ), as desired.

Proof Let X be a compact countably bisequential space, let ${\mathcal {B}}$ be a barrier, and let $f:{\mathcal {B}}\to X$ . Fix an ultrafilter ${\mathcal {U}}\subseteq {\mathcal {P}}(\omega )$ and define an ultrafilter ${\mathcal {V}}\subseteq {\mathcal {P}}(f[{\mathcal {B}}])\subseteq {\mathcal {P}}(X)$ by $V\in {\mathcal {V}}$ if and only if $f^{-1}(V)\in {\mathcal {U}}^{\mathcal {B}}$ .

As X is compact, we can find $x\in X$ such that $x\in \overline {\mathcal {V}}$ , and by bisequentiality, there exists a decreasing family $\{V_n:n\in \omega \}\subseteq {\mathcal {V}}$ such that for any open neighborhood ${W\in {\mathcal {N}}(x)}$ , there is $n\in \omega $ such that $V_n\subseteq W$ . Define $U_n=f^{-1}(V_n)$ for every $n\in \omega $ and let $T_n=\{s\in T({\mathcal {B}}):\exists b\in U_n(s\subseteq b)\}$ . Then, each $T_n$ is a ${\mathcal {U}}$ -branching tree.

We now recursively define an infinite set $M=\{m_i:i\in \omega \}$ . Let ${m_0=\min (\mathrm{succ}_{T_0}(\emptyset ))}$ . If we have already defined $\{m_i:i<n\}$ , define $m_n$ such that for every $s\subseteq \{m_i:i<n\}$ with $s\in T({\mathcal {B}})\setminus {\mathcal {B}}$ , we have that $s^\smallfrown m_n\in T_n$ . The choice of $m_n$ is possible since every relevant s is an element of $T_0$ and the trees $T_n$ are ${\mathcal {U}}$ -branching. Moreover, if $s\subseteq M$ , $s\in T({\mathcal {B}})\setminus {\mathcal {B}}$ and $\min (s)\geq m_n$ , then $s\in T_k$ for every $k\geq n$ .

It remains to show that $f\upharpoonright ({\mathcal {B}}|M)$ converges to x. Fix an open set $W\in {\mathcal {N}}(x)$ and find $n\in \omega $ such that $V_n\subseteq W$ . Given $s\in {\mathcal {B}}|(M\setminus m_n)$ , let $m_k=\max (s)$ and ${s'=s\setminus \{m_k\}}$ . By the choice of $m_k$ , we have that $s= {s'}^\smallfrown m_k \in T_n\cap {\mathcal {B}}=U_n$ as $k\geq n$ ; thus, ${f(s)\in V_n\subseteq W}$ . Therefore, $f[{\mathcal {B}}|(M\setminus m_n)]\subseteq W$ .

Proof Note that a compact space is angelic if and only if it is Fréchet. In [Reference Corral, Guzmán and López-Callejas7], an almost disjoint family ${\mathcal {A}}\subseteq [\omega \times \omega ]^\omega $ , such that ${\mathcal {F}}({\mathcal {A}})$ is Fréchet, sequentially compact, but fails to be 2-sequentially compact, is constructed. Hence, ${\mathcal {F}}({\mathcal {A}})$ is an angelic space.

In order to prove that ${\mathcal {F}}({\mathcal {A}})$ is not 2-sequentially compact, ${\mathcal {A}}$ was constructed so that $A_n:=\{(n,m):m\in \omega \}\in {\mathcal {A}}$ for every $n\in \omega $ and so that $G:[\omega ]^2\to \omega \times \omega $ , defined by $G(\{n,m\})=(n,m)$ for $n<m$ , does not converge to any point in ${\mathcal {F}}({\mathcal {A}})$ . To avoid confusion, let’s denote by $\ast $ the point at infinity on ${\mathcal {F}}({\mathcal {A}})$ . As $\lim _{m\to \infty }(n,m)=A_n$ and any infinite subset of ${\mathcal {A}}$ converges to $\ast $ in ${\mathcal {F}}({\mathcal {A}})$ , we get that ${\lim _{n\to \infty }\lim _{m\to \infty }(n,m)=\ast} $ . Then, moreover, we have that ${\mathcal {F}}({\mathcal {A}})$ does not have the Ramsey property.

Question 5.1 Is there a compact and $\omega _1$ -sequentially compact space that is not countably bisequential?

Proof We prove it by induction on $\rho ({\mathcal {B}})$ . If ${\mathcal {B}}=[\omega ]^1$ and $X\in \textsf {FIN}^{\mathcal {B}}=\textsf {FIN}$ , define $H\subseteq \omega ^{<\omega }$ such that $\emptyset \in H$ , $\mathrm{succ}_H(\emptyset )=\omega \setminus k$ , where $X\subseteq k$ and $\mathrm{succ}_H(s)=\omega $ for any $s\in H\setminus \{\emptyset \}$ . This H clearly works.

Let now ${\mathcal {B}}$ be an arbitrary barrier with $\rho ({\mathcal {B}})>1$ . We may assume that $\{k\}\notin {\mathcal {B}}$ for every $k\in \omega $ . Fix $X\in \textsf {FIN}^{\mathcal {B}}$ . Thus, $\{n\in \omega :X\cap {\mathcal {B}}(n)\notin \textsf {FIN}^{{\mathcal {B}}{[n]}}\}$ is finite, and we can find $k\in \omega $ that contains this set. Hence, for every $n\in \omega \setminus k$ , there is a Hechler tree $H_n$ with root $\{n\}$ such that $X\cap H_n\cap {\mathcal {B}}(n)=\emptyset $ . Define $H=\{\emptyset \}\cup \bigcup _{n\geq k}H_n$ . It is clear that H is a Hechler tree and $X\cap H\cap {\mathcal {B}}=\emptyset $ .

Proof Let ${\mathcal {X}}\subseteq \textsf {FIN}^{\mathcal {B}}$ with $|{\mathcal {X}}|<{\mathfrak {b}}$ . By Lemma 5.4, for every $X\in {\mathcal {X}}$ , there is a Hechler tree $H_X$ such that $X\cap H_X\cap {\mathcal {B}}=\emptyset $ . Let $f_X=f_{H_X}$ for every $X\in {\mathcal {X}}$ . Since $|{\mathcal {X}}|<{\mathfrak {b}}$ , we can find $f\in \omega ^\omega $ such that $f>^*f_X$ for every $X\in {\mathcal {X}}$ . It follows from our enumeration of $\omega ^{<\omega }$ that if $f>^*f_X$ , then there exists $n_X\in \omega $ such that if $f_X(s)>f(s)$ for some $s\neq \emptyset $ , then $s\in {\mathcal {B}}(i)$ for some $i\leq n_X$ . Thus, letting $H=H_f$ , we have that $H\setminus H_X\subseteq \bigcup _{i\leq n_X}{\mathcal {B}}(i)$ , and this implies that

$$ \begin{align*} X\cap H\cap{\mathcal{B}}\subseteq\bigcup_{i\leq n_X}{\mathcal{B}}(i).\\[-42pt]\end{align*} $$

Proof As ${\mathcal {C}}$ is nontrivial, we may assume without loss of generality that every element of the ${\mathcal {C}}$ has size at least $2$ ; in particular, ${\mathcal {C}}(m)$ is infinite for every $m\in \omega $ .

Consider $i:{\mathcal {C}}\to {\mathcal {C}}$ to be the identity map and $E\in [\omega ]^\omega $ . We shall show that ${i\upharpoonright ({\mathcal {C}}|E)}$ does not converge in ${\mathcal {F}}({\mathcal {A}})$ . Since $i[{\mathcal {C}}|E]$ is infinite, it does not converge to any isolated point in ${\mathcal {C}}$ .

However, if $A\in {\mathcal {A}}$ and $n\in \omega $ , we have that $|{\mathcal {C}}(m)\cap i[{\mathcal {C}}|E\setminus n]|=\omega $ for every ${m\in E\setminus n}$ , but $|A\cap {\mathcal {C}}(m)|\leq 1$ implies that $i[{\mathcal {C}}|(E\setminus n)]\nsubseteq A$ , and hence, it does not converge to A. It remains to show that it does not converge to $\infty $ .

Fix the neighborhood $U={\mathcal {F}}({\mathcal {A}})\setminus (\{A\}\cup A)$ of $\infty $ for some $A\subseteq {\mathcal {C}}|E$ and let $n\in \omega $ . As $|A\cap {\mathcal {C}}(i)|\leq 1$ for every $i\in \omega $ , we can find $c\in A\subseteq {\mathcal {C}}|E$ such that $\min (c)>n$ . Thus, $c\in {\mathcal {C}}|(E\setminus n)\setminus U$ , and in consequence, $i[{\mathcal {C}}|(E\setminus n)]\nsubseteq U$ . Since this is true for any $n\in \omega $ , we get that $i\upharpoonright ({\mathcal {C}}|E)$ does not converge to $\infty $ .

Proof To see that (1) implies (2), let $f:Y\rightarrow X$ . As ${\mathcal {I}}\not \leq _K{\mathcal {J}}$ , f is not a Katětov function, so there is $A\in {\mathcal {I}}$ such that $B:=f^{-1}(A)\in {\mathcal {J}}^+$ . Now $f[B]\in {\mathcal {I}}$ since $f[B]\subseteq A\in {\mathcal {I}}$ . Conversely, to see that ${\mathcal {I}}\not \leq _K{\mathcal {J}}$ , let $f:Y\rightarrow X$ ; we want to prove that f is not a Katětov function. So let $B\in {\mathcal {J}}^+$ such that $A:=f[B]\in {\mathcal {I}}$ . Now $f^{-1}(A)\in {\mathcal {J}}^+$ since $B\subseteq f^{-1}(A)$ .

Proof Enumerate ${\mathcal {C}}^{\mathcal {B}}=\{f_\alpha :\alpha <{\mathfrak {c}}\}$ and $[\omega ]^\omega =\{E_\alpha :\alpha <{\mathfrak {c}}\}$ . For $E\in [\omega ]^\omega $ , let $E^\uparrow =T({\mathcal {C}}|E)$ . Notice that for any such E, the tree $E^\uparrow $ is everywhere $\omega $ -splitting, and hence, it has infinite intersection with any Hechler tree H inside ${\mathcal {C}}$ ; that is, $|E^\uparrow \cap H\cap {\mathcal {C}}|=\omega $ . Moreover, for any $n\in E$ , we have that

$$ \begin{align*}|E^\uparrow\cap H\cap {\mathcal{C}}(n)|=\omega.\end{align*} $$

Recursively construct $\{A_\alpha :\alpha <{\mathfrak {c}}\}\subseteq [{\mathcal {C}}]^\omega $ and $\{X_\alpha :\alpha <{\mathfrak {c}}\}\subseteq [\omega ]^\omega $ such that for all $\beta <\alpha <{\mathfrak {c}}$ ,

1. (1) $f_\alpha [{\mathcal {B}}|X_\alpha ]\in \textsf {FIN}^{\mathcal {C}}$ ,

2. (2) $|A_\alpha \cap A_\beta |<\omega $ ,

3. (3) $|A_\alpha \cap f_\beta [{\mathcal {B}}|X_\beta ]|<\omega $ ,

4. (4) $|A_\alpha \cap {\mathcal {C}}(n)|\leq 1$ for all $n\in \omega $ ,

5. (5) $A_\alpha \subseteq E_\alpha ^\uparrow $ .

At step $\alpha <{\mathfrak {c}}$ , let $X_\alpha \in [\omega ]^\omega $ such that $f_\alpha [{\mathcal {B}}|X_\alpha ]\in {\mathsf {FIN}}^{\mathcal {C}}$ (this set exists by Lemma 5.9). Note that also every $A_\beta $ with $\beta <\alpha $ is an element of $\textsf {FIN}^{\mathcal {C}}$ . Then, we can find a Hechler tree H as in Lemma 5.5 such that

$$ \begin{align*}X\cap H\cap{\mathcal{C}}\subseteq \bigcup_{i\leq n_X}{\mathcal{C}}(i)\end{align*} $$

for either $X=f_\beta [{\mathcal {B}}|X_\beta ]$ or $X=A_\beta $ for some $\beta <\alpha $ . Let $A_\alpha \in [{\mathcal {C}}]^\omega $ be such that $A_\alpha \subseteq H\cap {\mathcal {C}}\cap E_\alpha ^\uparrow $ and $|A_\alpha \cap {\mathcal {C}}(i)|\leq 1$ for every $i\in \omega $ . It follows that $A_\alpha \cap X\subseteq \bigcup _{i\leq n_X}{\mathcal {C}}(i)$ for some $n_X\in \omega $ , and as $|A_\alpha \cap {\mathcal {C}}(i)|\leq 1$ for every $i\leq n_X$ , this intersection is finite, and (2) and (3) hold. The remaining three properties follow from the construction.

Since $\{A\in {\mathcal {A}}:|f_\alpha [{\mathcal {B}}|X_\alpha ]\cap A|=\omega \}\subseteq \{A_\beta :\beta \leq \alpha \}$ , and this set has size less than ${\mathfrak {b}}$ , by Proposition 5.6, we get that ${\mathcal {F}}({\mathcal {A}})$ is ${\mathcal {B}}$ -sequentially compact.

However, item (5) implies that ${\mathcal {F}}({\mathcal {A}})$ is not ${\mathcal {C}}$ -sequentially compact by Lemma 5.7, as $\mathcal C$ has rank greater than 1.

Proof By Lemma 5.9, it is enough to prove that for every $f:{\mathcal {B}}\to {\mathcal {C}}$ , there is $X\in [\omega ]^\omega $ such that $f[{\mathcal {B}}|X]\in {\mathsf {FIN}}^{\mathcal {C}}$ . The proof is by induction on $\rho ({\mathcal {C}})$ . If $\rho ({\mathcal {C}})=1$ , there is nothing to do, so we can assume that $\rho ({\mathcal {C}})>1$ , and the result is true for ${\mathcal {C}}'$ whenever $\rho ({\mathcal {C}}')<\rho ({\mathcal {C}})$ . We now prove it by induction on $\rho ({\mathcal {B}})$ . If $\rho ({\mathcal {B}})=1$ , we can find an infinite set $X\in [\omega ]^\omega $ such that either $|f[{\mathcal {B}}|X]\cap {\mathcal {C}}(i)|\leq 1$ for every $i\in \omega $ or $f[{\mathcal {B}}|X]\subseteq {\mathcal {C}}(n)$ for some $n\in \omega $ . Hence, assume that for any barriers ${\mathcal {B}}'$ and ${\mathcal {C}}'$ with $\rho ({\mathcal {B}}')<\rho ({\mathcal {B}})$ , $\rho ({\mathcal {C}}')\leq \rho ({\mathcal {C}})$ , and $\rho ({\mathcal {B}}')<\rho ({\mathcal {C}}')$ , the result holds. We can also think of $\{\emptyset \}$ as a barrier of rank 0. In this case, the result is trivial, but it is worth mentioning since ${\mathcal {B}}[s]$ is a barrier of rank 0 whenever $s\in {\mathcal {B}}$ . We will also consider these kind of barriers as part of our inductive hypothesis. We can also assume, by possibly passing to a cofinite set, that $\rho ({\mathcal {C}}(n))\geq \rho ({\mathcal {B}})$ for every $n\in \omega $ .

Define a function $\pi _n:{\mathcal {B}}\to 2$ for every $n\in \omega $ given by $\pi _n(s)=0$ if and only if $f(s)\in \bigcup _{i\leq n}{\mathcal {C}}(i)$ . By Nash-Williams, we can find an infinite set $Z_n\in [\omega ]^\omega $ such that $\pi _n\upharpoonright ({\mathcal {B}}|Z_n)$ is constant. If $\pi _n\upharpoonright ({\mathcal {B}}|Z_n)$ is constant with value i, we will say that $Z_n$ is a monochromatic set with color i. If for some $n\in \omega $ we can find a monochromatic set $Z_n$ with color $0$ , then f maps ${\mathcal {B}}|Z_n$ into the union of the first $n+1$ -cones of ${\mathcal {C}}$ , which is an element of $\textsf {FIN}({\mathcal {C}})$ . Hence, we can assume that there is no monochromatic set with color $0$ for every $n\in \omega $ .