Proof. Consider the following infinite two-player game $G(\mathcal F, {\mathcal I})$ :

I winning if $\{j_n: n\in \omega \}\in {\mathcal I}$ .

It is standard and easy to see that Player I has a winning strategy if and only if (1) holds and Player II has a winning strategy if and only if (2) holds.

For (1) this is an immediate. For (2) it should be clear that any $\mathcal F^+$ -branching tree permits player II to play along a branch of the tree (the branching sets being in $\mathcal F^+$ intersect every potential move of Player I, hence, permit Player II to continue playing along a branch of the tree). If, moreover, all branches of the tree are in the ideal $\mathcal I$ then the strategy of playing along a branch of the tree becomes automatically winning.

The construction of the tree from a winning strategy has a tiny trick. The natural approach is to collect all the answers of player II to partial plays of the game building in this way an $\mathcal F^+$ -branching tree. The problem, however, is that such a tree may be too wide, i.e., possibly not every branch of such a tree corresponds to (the answers of Player II) a full play of the game according to the (winning) strategy. To overcome the problem, we can prune the tree having Player II fix for every node of the pruned tree a partial play of the game played according to the strategy in a coherent way (that is a longer node corresponds to a longer partial play of the game). In this way we still get an $\mathcal F^+$ -branching tree, but now, as all branches come from plays of the game played according to the winning strategy, they are all in $\mathcal I^+$ .

The determinacy of the game then provides the proof of the dichotomy.

Proof. Consider the Definable Ideal Dichotomy for $\mathcal F={\mathcal J}^*$ and ${\mathcal I}$ —the ideal dual to the neighborhood filter of x.

If there is an $\mathcal F^+$ -branching tree T with all branches in ${\mathcal I}^+$ then $\{{\mathrm {succ}}_T(t): t \in T\}$ forms a local $\pi $ -network Footnote ⁵ at x. To see this, assume toward a contradiction that it is not true, i.e., there is a neighborhood U of x which does not contain ${\mathrm {succ}}_T(t)$ for any $ t \in T$ . This means that is an element of ${\mathcal I}$ such that ${\mathrm {succ}}_T(t)\cap I\neq \varnothing $ for any $t\in T$ hence one can recursively construct a branch of T contained in I which is a contradiction.

If there is an $\mathcal F$ -branching tree with all branches in ${\mathcal I}$ then captures convergent sequences. Again, assume not, i.e., there is a sequence C convergent to x intersecting ${\mathrm {succ}}_T(t)$ for every $t \in T$ , so one can construct a subsequence of C which is a branch of T hence is in ${\mathcal I}^\perp $ and, in particular, in ${\mathcal I}^+$ , contradiction.

Proof. It is enough to show that whenever $({\mathbb G},\tau )$ is not metrizable and $({\mathbb G},{[\tau ]})$ is not $k_\omega $ , $({\mathbb G},\tau )$ is not strongly groomed. To that end, consider the ideal ${\mathcal I}$ generated by closed locally compact subspaces of $({\mathbb G},\tau )$ . Apply Theorem 11 and observe that alternative (1) implies that $({\mathbb G},{[\tau ]})$ is $k_\omega $ contradicting our assumption. Thus alternative (2) holds and ${\mathbb G}$ has a countable local $\pi $ -network ${\mathcal H}$ at ${1_{{\mathbb G}}}$ that consists of non-locally compact closed subsets: The alternative (2) provides a local $\pi $ -network ${\mathcal H}$ at ${1_{{\mathbb G}}}$ consisting of sets whose closures are not locally compact. Regularity of the group topology implies that the closures of elements of ${\mathcal H}$ also form a local $\pi $ -network at ${1_{{\mathbb G}}}$ as every neighborhood U of ${1_{{\mathbb G}}}$ contains the closure of a neighborhood V of ${1_{{\mathbb G}}}$ which in turn contains some $H\in {\mathcal H}$ , hence $\overline {H}\subseteq U$ .

Applying Lemma 12 and Theorem 11 to the ideal $ \mathbf{nwd}({[\tau ]})$ we conclude that if $({\mathbb G},\tau )$ is strongly groomed, it has a countable local $\pi $ -network at ${1_{{\mathbb G}}}$ that consists of ${[\tau ]}$ -open subsets. Now Lemma 13 implies that $({\mathbb G},\tau )$ is not strongly groomed.

Proof. Consider the ideal ${\mathcal I}$ generated by sequences that converge to x and apply Theorem 11. Alternative (1) would mean there exists a countable family of convergent sequences such that every convergent sequence in X is covered by finitely many members of the family. Standard arguments show that if such a family can be chosen to be finite then (1) holds, otherwise (2) follows.

Now, alternative (2) of Theorem 11 implies (3).

Proof. Consider the DID for $\mathcal F=\mathcal F^+=\mathcal U$ and a Borel ideal ${\mathcal I}$ .

If there is an $\mathcal F^+$ -branching tree T with all branches in ${\mathcal I}$ then $\mathcal U\cap {\mathcal I}\neq \emptyset $ . Otherwise there is an $\mathcal F$ -branching tree with all branches in ${\mathcal I}^+$ which leads to a contradiction.

Proof. The construction will give a topology on the ordinal $\omega ^\omega +1$ where the topology differs from the compact order topology only in the neighborhoods of the last point where the new neighborhoods will come from a filter $\mathcal F$ of clopen subsets of the ordinal $\omega ^\omega $ with the property that for every sequence $C\subseteq \omega ^\omega $ order converging to the point $\omega ^\omega $ there is an $F\in \mathcal F$ disjoint from C.

As the filter consists of clopen subsets of $\omega ^\omega $ the topology below $\omega ^\omega $ does not change, and as no old sequence from $\omega ^\omega $ converges to $\omega ^\omega $ , the space is not sequential. On the other hand, every subset of the space which is not closed discrete will contain a convergent sequence, as the original space was compact.

To construct the filter $\mathcal F$ we shall introduce the following notation. For every ordinal $\alpha <\omega ^\omega $ let $I_\alpha $ be the largest clopen interval with maximum $\alpha $ which is order isomorphic to an indecomposable ordinal. Note that:

1. (1) for each $\alpha , \beta <\omega ^\omega I_\alpha \subseteq I_\beta $ , $I_\beta \subseteq I_\alpha $ , or $I_\alpha \cap I_\beta =\emptyset $ ,

2. (2) each $I_\alpha $ is order isomorphic to $\omega ^n$ for some $n\in \omega $ , where $\omega ^0$ is a singleton,

3. (3) minimum of each $I_\alpha $ is either $0$ or the successor of an ordinal $\beta $ such that $I_\beta $ and $I_\alpha $ are order isomorphic, and

4. (4) each $I_\alpha $ of order type $\omega ^{n+1}$ is a disjoint union of an increasing $\omega $ -sequence of $I_{\beta _n}$ each of order type $\omega ^n$ with a last limit point on top.

For each $\alpha <\omega ^\omega $ we shall define a family $\mathcal F_\alpha $ of subsets of $I_\alpha $ using recursion on the order type of $I_\alpha $ as follows:

1. (1) If $I_\alpha $ is a singleton $\{\alpha \}$ then $\mathcal F_\alpha =\{\emptyset \}$ , otherwise.

2. (2) $I_\alpha $ is of order type $\omega ^{n+1}$ and is a disjoint union of an increasing $\omega $ -sequence of $I_{\beta _k}$ of order type $\omega ^n$ . Then

   $\mathcal F_\alpha =\{ A\subseteq I_\alpha : \ \exists K\in \omega \text { such that } I_{\beta _K}\subseteq A \text { while } $

   $ A\cap I_{\beta _k}=\emptyset \text { for } k>K \text { and } A\cap I_{\beta _k}\in \mathcal F_{\beta _k} \text { for } k<K \}.$

Let

$$ \begin{align*}\mathcal F=\{ A\subseteq \omega^\omega: \ \forall n\in \omega \ A\cap \omega^n \in \mathcal F_{\omega_n}\}.\end{align*} $$

The family $\mathcal F$ is a centered family of clopen subsets of $\omega ^\omega $ such that for all sequences $C\subseteq \omega ^\omega $ convergent to $\omega ^\omega $ there is an $A\in \mathcal F$ such that $A\cap C=\emptyset $ . To define the space X let $\mathcal F$ be the basis of open neighborhoods of $\omega ^\omega $ .

Question 1. Is there a locally compact remotely sequential non-sequential space that does not contain a copy of X?

Proof. We will show that ${\mathbb B}(X)$ and ${\mathbb B}(X,{[\tau ]})$ share the same compact and closed discrete subspaces where $\tau $ is the topology on X constructed in Example 20 and ${[\tau ]}$ is the sequential coreflection of $\tau $ . Note that ${\mathbb B}(X,{[\tau ]})$ is $k_\omega $ since $(X,{[\tau ]})$ is a topological sum of countably many compact subspaces.

Let $X=\bigcup _{i\in \omega }B_i\cup \{\infty \}$ be the remotely sequential non-sequential space constructed in Example 20. Here $B_i$ ’s are compact and disjoint with the property that $A\subseteq X$ is closed and discrete in X provided every intersection $A\cap B_i$ is finite. Additionally $\infty \in \overline {\bigcup _{i\in \omega }B_i}$ , which makes X a non-sequential space.

Consider the free Boolean topological group ${\mathbb B}(X,\tau )$ over X. Given an infinite convergent sequence $S\subseteq {\mathbb B}(X)$ , let $\tau _0$ be the coarser compact topology on X in which $\bigcup _{i\in \omega }B_i\cup \{\infty \}$ is a one-point compactification of the topological sum $\bigcup _{i\in \omega }B_i$ . Then S remains convergent in ${\mathbb B}(X,\tau _0)$ . Thus there exists an $n\in \omega $ such that $S\subseteq \sum ^nX$ . Without loss of generality, passing to a subsequence if necessary we may assume that for some infinite $S'\subseteq S$ , where $a_j^i$ are distinct for any given $i\in \omega $ . By passing to a subsequence again we may further assume that $a_j^i\in \cup _{s<k}B_s$ for all $j\leq m\leq n$ , $i\in \omega $ , and some $k\in \omega $ , and $a_j^i\in B_{m(i,j)}$ where ${\langle m(i,j):i\in \omega \rangle }$ is strictly increasing for each $j>m$ (this is only possible if $m<n$ ).

Suppose $m<n$ . Then the set is closed and discrete in X. Let be a clopen subset of X such that $T\subseteq F$ . Let $\tau _1$ be the topology on the topological sum in which F inherits its topology from X, and is a one-point compactification of . Now $S'$ is not convergent in ${\mathbb B}(X,\tau _1)$ , and hence in ${\mathbb B}(X)$ . Therefore $m=n$ and $S'\subseteq \cup _{s<k}B_s$ for some $k\in \omega $ and an infinite $S'\subseteq S$ .

A standard argument using the property that ${\mathbb B}(X,{[\tau ]})$ is $k_\omega $ shows that S is a convergent sequence in ${\mathbb B}(X,{[\tau ]})$ where ${[\tau ]}$ is the topology of the topological sum $\bigcup _{i\in \omega }B_i\cup \{\infty \}$ as well as the sequential coreflection of the original topology on X. Hence the sequential coreflection of ${\mathbb B}(X)$ is ${\mathbb B}(X,{[\tau ]})$ .

Let $L\subseteq {\mathbb B}(X,{[\tau ]})$ be a closed discrete subset. Put $L_i=L\cap \sum ^iX$ . Without loss of generality, we may assume that every $x\in L_k$ has the form $x=a_1+\cdots +a_k$ where $a_j\not =\infty $ . Put and . Finally, define for each $k\in \omega $ . Note that $F_k\cap B_i$ is finite for every $k,i\in \omega $ . Indeed, otherwise, for some $k,r\in \omega $ , there would be infinitely many $x\in L_k$ such that $m(x)=r$ . It then follows from the definition of $m(x)$ that $L_k\cap \bigcup _{j\leq r}B_j$ is infinite, contradicting that $L\supseteq L_k$ is closed and discrete in ${\mathbb B}(X,\tau _2)$ . Note that .

Define $F=\bigcup _{k\in \omega }F_k\cap (\bigcup _{i\geq k}B_i)$ . Observe that each $F\cap B_i$ is finite and find a $U\subseteq \bigcup _{i\in \omega }B_i$ , clopen in X such that $F\subseteq U$ .

Consider the intersection . Suppose $L'$ is infinite. Then $m(x)$ for $x\in L'$ is unbounded so there exists an $x\in L'$ such that $m(x)>\max \{k, m+n\}$ . For such x, $t(x)\subseteq F_{m+n}$ , $t(x)\cap B_k=\varnothing $ , and, since $m(x)>m+n$ , $t(x)\subseteq F\subseteq U$ . This implies . This is a contradiction since $x=\sum x'+\sum t(x)$ where $x'\subseteq X$ , $x'\cap t(x)=\varnothing $ , and this representation is unique.

Consider the topology $\tau _U$ on X in which X is a topological sum of U and , where U retains its inherited topology while is a one-point compactification of . Note that $\tau _U$ is coarser than the original topology of X and makes X a topological sum of countably many compact subspaces . The argument in the previous paragraph implies that L is a closed discrete subspace of the $k_\omega $ group ${\mathbb B}(X,\tau _U)$ and hence of ${\mathbb B}(X)$ .

This shows that ${\mathbb B}(X)$ and ${\mathbb B}(X,{[\tau ]})$ share compact as well as closed discrete subspaces.

Construction. Let $X=\{\omega \}\cup \omega \cup \omega \times \omega \cup \omega ^\omega $ and let be a 1-1 listing of some AD family on $\omega $ .

Define the topology on X as follows. All the points in $\omega \times \omega $ are isolated, basic neighborhoods of $(m,n)\in \omega \times \omega $ are , $k\in \omega $ . For each $f\in \omega ^\omega $ let $s_f\subseteq \omega \times \omega $ be defined as . Let, as a subspace of X, $\{\omega \}\cup \omega ^\omega $ be the one-point compactification of the infinite discrete subspace $\omega ^\omega $ . Declare $c_f\to f$ and $s_f\to \omega $ to be convergent sequences for every $f\in \omega ^\omega $ .

Let $\tau $ be the finest topology on X for which all of the conditions above hold. Since everything was stated in terms of convergent sequences, $(X,\tau )$ is sequential. Let $U\ni \omega $ be open in $\tau $ . Then for some finite $F\subseteq \omega ^\omega $ .

Define $U'=\{\omega \}\cup (((U\cap \omega )\times \omega )\cap U)\cup \bigcup _{f\in F}(s_f\cap U)\cup (U\cap \omega ^\omega )\subseteq U$ . Thus $U'$ is U modified above finitely many $c\in {\mathcal A}$ . Let be an infinite sequence. If $S\subseteq \omega \times \omega $ then we may assume that $S\subseteq s_f$ for some $f\in \omega ^\omega $ or that $S\subseteq \{n\}\times \omega $ for some $n\in \omega $ . In the former case, if $f\not \in F$ then $s_f\subseteq ^*((U\cap \omega )\times \omega )\cap U\subseteq U'$ since U is open and $\omega \in U$ , contradicting . Thus $f\in F$ and $s_f\subseteq ^* s_f\cap U$ , again, contradicting .

If $S\subseteq \{n\}\times \omega $ and $n\in U'\subseteq U$ then $S\subseteq ^* U'$ by the construction of $U'$ , a contradiction.

Suppose $S\subseteq \omega $ . Then we may assume that $S\cap c_f$ is infinite for some $f\in \omega ^\omega $ . If $f\in U$ then $f\not \in F$ and $S\cap U$ is infinite, a contradiction. Thus $f\in F$ so $S\to f\not \in U'$ .

The case when $S\subseteq \omega ^\omega $ is trivial. Since any converging sequence contains an infinite subsequence of one of the kinds treated above, we have shown that $U'$ is open. A similar argument shows that $U'$ is closed. Since it is clear that every other point of X has a base of clopen neighborhoods, $(X,\tau )$ is regular. Thus $(Y,\tau )$ is also regular as a subspace of X where $Y=\{\omega \}\cup \omega \times \omega \cup \omega $ . Note that, by the construction of X there are no $S\subseteq \omega $ such that $S\to \omega $ thus making $\omega \subseteq Y$ closed and discrete in $(Y,{[\tau ]})$ .

Let $V\ni \omega $ be open in ${[\tau ]}$ . Suppose each $\{n\}\times \omega \cap V$ is finite and define $g:\omega \to \omega $ as follows: if $\{n\}\times \omega \cap V\not =\varnothing $ let ; otherwise let $g(n)=0$ . Now the choice of $s_g$ ensures that contradicting the assumption that V is open in ${[\tau ]}$ and $\omega \in V$ (recall $s_g\to \omega $ in $\tau $ and thus in ${[\tau ]}$ ).

The argument above shows that there is an $n\in \omega $ such that $n\in \overline {V}^{{[\tau ]}}$ , which means that $\omega \in Y$ and $\omega \subseteq Y$ cannot be separated by open neighborhoods implying $(Y,{[\tau ]})$ is not regular.

Note that in the argument above, all that is needed for the construction of X is a 1-1 correspondence between $\omega ^\omega $ and some AD family on $\omega $ . Since such correspondence can easily be made analytic we can see that Y can be made analytic as well.

Construction. Let ${\mathbb G}={\mathbb B}(Y)$ be the free Boolean group over the space Y constructed in the example above. Note that Y is a closed subspace of $({\mathbb G},\tau )$ and thus of $({\mathbb G},{[\tau ]})$ where $\tau $ is the free Boolean group topology on ${\mathbb G}$ . Thus if $U\subseteq Y$ is open in $(Y,{[\tau ']})$ where $\tau '$ is the original topology on Y (which is also the topology Y inherits from ${\mathbb G}$ ) then is closed in $({\mathbb G},{[\tau ]})$ . This shows that the topology Y inherits from $({\mathbb G},{[\tau ]})$ is ${[\tau ']}$ implying that $({\mathbb G},{[\tau ]})$ is not regular.

Proof. Let $\mathcal I$ be the $F_\sigma $ -ideal generated by branches of the Cantor tree $2^{<\omega }$ and let $\mathcal J$ be the ideal orthogonal to the ideal generated by an $\aleph _1$ -dense set of branches, and let $\mathcal G$ be the filter dual to $\mathcal J$ and let $X=\omega \cup \{\mathcal F\}$ be the space with a unique non-isolated point $\mathcal F$ . Then $\mathsf {IA}(X,\mathcal I)$ fails.

The fact that (1) fails follows immediately from the definition, the $\aleph _1$ -many branches are convergent sequences, and to see that (2) fails one needs to show that for any countable family of infinite antichains there is an element of $\mathcal J$ intersecting them all in an infinite set. By the disjoint refinement lemma we can assume that the antichains $\{a_n: n \in \omega \}$ are pairwise disjoint as the set of branches is of size $\aleph _1 < \mathfrak b$ there are finite subsets $F_n$ of $A_n$ such that is almost disjoint from all the $\aleph _1$ many branches hence is an element of $\mathcal J$ and intersects all of the $A_n$ in an infinite set.

Question 2. Can such an example be constructed in ZFC?

Proof. Consider the free Boolean group ${\mathbb G}={\mathbb B}(X)$ over X. Let ${\mathcal J}$ be the ideal in ${\mathbb G}$ generated by . Note that ${{\mathcal J}}|_{ }X={\mathcal I}$ so ${\mathcal J}$ does not have the sequence capturing property (at any $g\in {\mathbb G}$ ) by the choice of ${\mathcal I}$ .

Suppose there exists a countable almost $\pi $ -network ${\mathcal {U}}\subseteq {\mathcal J}^+$ at $x_0\in X\subseteq G$ . Define $\pi :{\mathcal P}({\mathbb G})\to {\mathcal P}(X)$ by Note that $\pi $ is well defined since $X\subset {\mathbb G}$ forms a basis of ${\mathbb G}$ as a vector space over ${\mathbb F}_2$ . Put . Then ${\mathcal V}\subseteq {\mathcal I}^+$ is countable.

Suppose first that the family

is infinite. Let ${\mathcal {U}}^{\prime }= \langle U_i : i \in \omega \rangle $ where each $U\in {\mathcal {U}}^{\prime }$ is listed infinitely often, and recursively pick $\lambda _i\in [X]^{<\omega }$ such that $|\lambda _i|$ is strictly increasing, $\sum \lambda _i\in U_i$ . The set is closed and discrete in ${\mathbb G}$ . We may assume that $x_0\not \in D$ so there exists an open $V\subseteq {\mathbb G}$ such that $x_0\in V$ and $V\cap D=\varnothing $ . Now where $D\cap U$ is infinite. Thus is infinite so .

The argument above shows that we may assume that $U\subseteq \sum ^nX$ for every $U\in {\mathcal {U}}$ and some $n\in \omega $ .

Since ${\mathcal I}$ violates $\mathsf {IA}(\{X\},\cdot )$ at $x_0$ , there exists an open neighborhood $U\subseteq X$ of $x_0$ such that for every $V\in {\mathcal V}$ . Since $x_0$ is a unique non-isolated point of X, the subgroup ${\mathbb G}'={\langle {U}\rangle }\subseteq {\mathbb G}$ , algebraically generated by U, is open in ${\mathbb G}$ . Now for every $O\in {\mathcal {U}}$ thus ${\mathcal {U}}$ is not an almost $\pi $ -network of ${\mathcal J}^+$ subsets at $x_0$ .

Question 3. If ‘definable’ is dropped from the statement of Theorem 14 is the resulting statement consistent with ZFC? Does it follow from the IIA (see [Reference Hrušák and Shibakov10])?

Question 4. Let ${\mathbb G}$ be a countable (definable) group such that there exists a dense subset, which almost disjoint from every convergent sequence in G but every dense subgroup of ${\mathbb G}$ contains a convergent sequence. Can ${\mathbb G}$ be sequentially complete?