References on filter quantifiers

Solution 1:

We can generalize the usual notion of convergence by simply replacing the Fréchet filter $\mathbb F$ with any other filter on $\mathbb N$. I am interested in any references about this generalization.

My answer concerns one specific application of this convergence in general topology and topological algebra. Namely, given an (ultra)filter $\mathcal F$ on $\Bbb N$, a topological space $X$ (remark that we don't assume any separation axioms in our consideration of the definitions), and a sequence $\{x_n\}_{n\in\Bbb N}$ of points of $X$, we say that a point $x\in X$ is a $\mathcal F$-limit of $\{x_n\}$ provided for any neighborhood $U$ of $x$ a set $\{n\in\Bbb N: x_n\in U\}$ belongs to $\mathcal F$. This notion is known for a long time and was considered, for instance, by Bernstein [Ber], Furstenberg [Fur, p. 179], and Akin [Aki, p. 5, 61].

By $\Bbb N^*$ we denote the set of all free ultrafilters on $\Bbb N$. In this topic I am mainly acquainted with $\mathcal F$-compact spaces for given $\mathcal F\in\Bbb N^*$, that is such spaces $X$ in which any sequence has a $\mathcal F$ limit [Vau, Def. 4.5]. These spaces are important by the following two reasons.

Stratification of compact-like spaces. In general topology are well-studied different classes of compact-like spaces and relations between them, see, for instance, basic [Eng, Chap. 3] and general works [DRRT], [Mat], [Vau], [Ste], [Lip]. The including relations between classes are often visually represented by arrow diagrams, see, [Mat, Diag. 3 at p.17], [DS, Diag. 1 at p. 58] (for Tychonoff spaces), [Ste, Diag. 3.6 at p. 611], and [GR, Diag. at p. 3].

Recall that a space is countably compact iff each its open countable cover of has a finite subcover iff each its infinite subset $A$ has an accumulation point $x$ (the latter means that each neighborhood of $x$ contains infinitely many points of the set $A$). Clearly, if a space $X$ is $\mathcal F$-compact for some $\mathcal F\in\Bbb N^*$ then $X$ is countably compact (see [Vau, Lemma 4.6] for $X$ which are $T_3$).

We may call a space $X$ ultracompact if X is $\mathcal F$-compact for each $\mathcal F\in \Bbb N^*$. But it can be shown that each $\omega$-bounded space is ultracompact, and, conversely, by [Theorem 4.9, Vau] each ultracompact $T_3$ space is $\omega$-bounded. Recall that a space is $\omega$-bounded, if each its countable subset has compact closure.

Thus for any utrafilter $\mathcal F\in\mathbb N^*$ a class of $\mathcal F$-compact spaces is intermediate between the classes of $\omega$-bounded and countably compact spaces. An other intermediate class between these two classes is a class of totally countably compact spaces, that is spaces in which any sequence contains a subsequence with compact closure. There exists a Frolík class $\mathcal C$ of spaces, which is intermediate between totally countably compact and $\mathcal F$-compact for Tychonoff spaces, for an arbitrary, but fixed ultrafilter $\mathcal F\in\Bbb N^*$ (see the class Definition 3.9 in [Vau] (in which, I guess is missed that $Y$ is countably compact), the class characterization in [Vau, Theorem 3.11], and the diagram at [Vau, p.572]).

Given an utrafilter $\mathcal F\in\mathbb N^*$, it is easy to check that a product of any family of $\mathcal F$-compact spaces is $\mathcal F$-compact. On the other hand, it is easy to show that if the $2^\frak c$-th power of a space is countably compact, then the space is $\mathcal F$-compact for some $\mathcal F\in\Bbb N^*$.

A background of these results is the investigation of productivity of compact-like spaces, motivated by the fundamental Tychonoff theorem, stating that a product of a family of compact spaces is compact. On the other hand, there are two countably compact spaces whose product is not feebly compact (see [Eng, the paragraph before Theorem 3.10.16]). Recall that a space $X$ is pseudocompact, if each continuous real-valued function on $X$ is bounded. Clearly, each countably compact space is pseudocompact. A subclass of countably compact spaces is constituted by sequentially compact spaces. Recall that a space is sequentially compact, if each sequence of its points has a convergent subsequence. The product of a countable family of sequentially compact spaces is sequentially compact [Eng, Theorem 3.10.35]. But already the Cantor cube $\{0,1\}^\frak c$ is not sequentially compact (see [Eng, the paragraph after Example 3.10.38]). On the other classes of compact-like spaces preserved by products, see [Vau, $\S$ 3-4] (especially Theorem 3.3, Proposition 3,4, Example 3.15, Theorem 4.7, and Example 4.15) and $\S$7 for the history, and [Ste, $\S$ 5].

The Wallace problem. A semigroup $(S,\cdot)$ is cancellative if for any $a,b\in S$ we have $a=b$, provided there exists $x\in S$ such that $ax=bx$ or $xa=ab$. A semigroup $(S,\cdot)$ endowed with a topology is called a topological semigroup, if the multiplication map $S\times S\to S$ is continuous. If, moreover, $S$ is a group then $S$ is called a paratopological group. A paratopological group $G$ such that the inverse map $x\mapsto x^{-1}$ is continuous is a topological group.

The proof of [Tom$_3$, Theorem 3.1 in a preprint] shows that for each $\mathcal F\in\Bbb N^*$ each $\mathcal F$-compact cancellative Hausdorff topological semigroup $S$ is group. That is, $S$ is a paratopological group. Since $S\times S$ is countably compact, it is a topological group, see [RR]. Tomita also noted that we can modify proof of Theorem 3.1 to show that each sequentially compact cancellative semigroup $S$ is a group. Bokalo and Guran [BG, Theorem 6] also showed that each sequentially compact Hausdorff cancellative topological semigroup is a topological group.

A background of these results is the following well-known in topological algebra and still open under axiomatic assumptions problem [Wal], [Com$_2$] posed by Wallace in 1953. He remarked that several authors proved that a Hausdorff compact cancellative topological semigroup is a topological group and asked: whether every countably compact cancellative topological semigroup $S$ a topological group?

In 1957 Ellis [Ell] showed that a locally compact regular group even with separately continuous operation is a topological group (for a paratopological group we can drop regularity, see Proposition in this answer). In 1972, Mukherjea and Tserpes [MT] showed that the answer is affirmative for first countable semigroups. In 1993 Grant [Gra$_2$] obtained affirmative answers for particular cases. He also mentioned that it was known that the answer is affirmative for $\omega$-bounded semigroups.

On the other hand, in 1996 Robbie and Svetlichny [RS] showed under CH that there is a counterexample for the Wallace Problem. Tomita [Tom$_3$] showed that there is a counterexample to the Wallace Problem under $ MA_{countable}$. (Recall that $MA_{countable}$ is Martin's Axiom restricted to countable posets. This axiom is equivalent to a strong form of the Baire Category Theorem: a circle $\Bbb T$ is not a union of less than continuum many closed nowhere dense sets).

Now we have a following framework for counterexamples to the Wallace problem. Namely, let TT be the following axiomatic assumption: there is an infinite torsion-free abelian countably compact topological group without non-trivial convergent sequences. The first example of such a group constructed by Tkachenko under the Continuum Hypothesis [Tka]. Later, the Continuum Hypothesis weakened to the Martin Axiom for $\sigma$-centered posets by Tomita in [Tom$_3$], for countable posets in [KTW], and finally to the existence continuum many incomparable selective ultrafilters in [MuT]. Yet, the problem of the existence of a countably compact group without convergent sequences in ZFC seems to be open, see [DiS].

The proof of [BDG, Lemma 6.4] implies that under TT there exists a Hausdorff group topology on a free abelian group $F$ generated by the set $\frak c$ such that for each countable infinite subset $M$ of the group $F$ there exists an element $\alpha \in \overline M\cap\frak c$ such that $M\subset\langle \alpha \rangle$. Then the free abelian semigroup generated by the set $\frak c$ is countably compact, but not a group. Also we can modify a topology on $F$ making it a Hausdorff countably compact paratopological non-topological group, see [Rav, Example 2]. Conversely, by [Rav, Proposition 8], each such a group contains an element $g$ such that the closure $\overline{\langle g\rangle}$ of a cyclic subgroup $\langle g\rangle$ generated by an element $g$ is not a group. I recall that Pfister [Pfi] showed that each regular (locally) countably compact paratopological group is a topological group. Banakh et al. in [BBGGR] proved a bit more stronger result: a subgroup of a locally countably compact $T_3$ topological semigroup with open left shifts is a topological group.

As far as I know, a question whether there exists under ZFC a cancellative Hausdorff topological semigroup $S$ such that $S$ (or $S\times S$) is countably compact, but $S$ is not a group, is still unanswered.

See the surveys [Com$_1$], [Com$_2$] and [Chr] for a discussion on the Wallace problem [GKO], [Gra$_1$], [Gra$_2$], [Hel], [MT], [RS], [Tom$_1$]-[Tom$_6$], [Wal], [Yur] for other results on it.

Situation with counterparts of the Wallace problem with another compact-like properties is the following.

For pseudocompact case there is a following relatively simple counterexample. Let $H=\prod_{\alpha\in A} H_\alpha$ be a Tychonoff product of an uncountable family of compact topological groups, such that for each $\alpha\in A$ a group $H_\alpha$ contains a non-periodic element $h_\alpha$. For instance, for each $\alpha$ we can take as $H_\alpha$ the circle group $\Bbb T$. Let $G=\{(g_\alpha)\in H$ such that $g_\alpha=e$ for all but countably many $\alpha\in A\}$, $S$ be a subsemigroup of $H$ generated by $G$, and $h=(h_\alpha)$. Since $h^n\not\in G$ for any natural $n$, $h^{-1}\not\in S$, thus $S$ is not a group. Since $S$ contains a dense countably compact subset $G$, it is pseudocompact. Also there are feebly compact Hausdorff paratopological non-topological groups, see, for instance [Rav, Example 3] or [ST, Theorems 1 and 2]. On the other hand, there are many conditions providing that a feebly compact paratopological group is a topological group, see, for instance, [Rav, Proposition 3].

Given a cardinal $\kappa$, a space is $\kappa$-compact if every its open cover of size $\kappa$ has a finite subcover. We recall that Kunen's Axiom (KA) assumes that there exists a ultrafilter $\mathcal F\in\Bbb N^*$ with a basis of size $\omega_1$. KA trivially follows from CH, but also KA $\&$ $\frak c=\aleph_2$ is consistent, see [Kun, Ex. VIIIA.10]. In [Tom$_3$] Tomita proved that under KA each Tychonoff $\omega_1$-compact cancellative topological semigroup is a group. On the other hand, he showed that under MA$_{countable}$ $\&$ $\operatorname{cf}\frak c>\omega_1$ a topological group $\Bbb T^{\frak c}$ contains an $\omega_1$-compact subsemigroup, which is not a group. Remark that MA$_{countable}$ also is consistent with $\frak c = \aleph_2$ [Tom$_3$]. Therefore the answer to the question for Tychonoff $\omega_1$-compact semigroups is independent on $\frak c = \aleph_2$.

References

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[BBGGR] T. Banakh, S. Bardyla, I. Guran, O. Gutik, A. Ravsky, Positive answers for Koch's problem in special cases, in prepartion.

[BDG] T. Banakh, S. Dimitrova, O. Gutik, Embedding the bicyclic semigroup into countably compact topological semigroups, Topology Appl. 157:18 (2010) 2803--2814,

[Ber] A. Bernstein, A new kind of compactness for topological spaces, Fund. Math. 66 (1970), 185-193.

[Com$_1$] W. W. Comfort, Topological groups, in K. Kunen and J. E. Vaughan (eds.), Handbook of set theoretic topology}, Elsevier, 1984, 1143-1263.

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