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Using standard definitions, the topological space $Y$ is sequential if for each nonclosed $A \subset Y$, there exists a convergent sequence $a_{1}$ , $a_{2}$,...$\rightarrow b$ so that $a_{n} \in A$ but $b \notin A$.

Working in the topological category TOP, we assume the group $G$ is a sequential space, inversion is continuous, and group multiplication is continuous with respect to the standard product topology on $G \times G$.

Must $G \times G$ be a sequential space?

A `yes' answer would provide a sharp dividing line between sequential topological groups in TOP, and sequential topological groups in SEQ\TOP.

(SEQ is a category in which the standard product topology of $G \times G$ is refined to ensure $G \times G$ is sequential).

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Possibly a dumb question: so there is an example of a sequential topological space $X$ (not a group) such that $X \times X$ is not sequential? What is it? – Nate Eldredge Oct 12 '13 at 3:06
Yes. If X is the fundamental group of the Hawaiian earring (endowed with the quotient topology inherited naturally from the space of based loops). This is arguably overkill but not gratuitously so. For simpler but natural examples, there exist planar continua with countable free fundamental group X so that, topologized with the quotient topology as mentioned, X x X fails to be sequential. Arguably the root pathology lies within the definition of standard product topology, and such phenomena makes the case for the relevance of category theory. – Paul Fabel Oct 12 '13 at 3:37
More generally, if $q:A\to B$ is a quotient map where $A$ is first countable and $q\times q$ is not a quotient map, then $B\times B$ cannot be sequential. Another example: let $A=\coprod_{n\geq 0}(\mathbb{Q}\sqcup \mathbb{Q}^{-1})^n$ be free topological monoid on two disjoint copies of the rationals $\mathbb{Q}$ (certainly metrizable), $B=F(\mathbb{Q})$ be the free group on the rationals, and $q$ send a word to the corresponding reduced word in $F(\mathbb{Q})$. – Jeremy Brazas Oct 12 '13 at 5:02
There is a survey by Moore and Todorcevic on the related question of whether every separable Frechet group is metrizable. At the end, they ask the question (credited to Hrusak) of whether this might hold under Todorcevic's Open Coloring Axiom. I'm not sure what the current state of this is. – Iian Smythe Oct 17 '13 at 1:07
up vote 10 down vote accepted

This is consistently false.

It was proved by Malyhin and Shakhmatov (see "Cartesian products of Frechet topological groups and function spaces", Acta Math. Hung. 1992) that in any model obtained by adding a Cohen real to a model of $MA + \lnot CH$, there is a sequential topological group $G$ such that $G \times G$ is not sequential. In fact they can get $G$ to be Frechet-Urysohn and hereditarily separable (which could not have been done in $ZFC$ since consistently every separable Frechet topological group is metrizable1) while $G \times G$ has uncountable tightness.

In $ZFC$ there are examples of sequential (in fact Frechet-Urysohn) topological groups $G$ and $H$ such that $G \times H$ is not sequential (in fact not countably tight). This was shown by Todorcevic in "Some applications of $S$ and $L$ combinatorics", Ann. New York Acad. Sci. 1993.

I don´t know if there is a $ZFC$ example of a sequential topological group with a non-sequential square.

1 See M. Hrušák and U.A. Ramos-García, Malykhin's problem. Adv. Math. 262 (2014), 193–212.

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This is consistently true, for separable Frechet $G$.

In a recent preprint, Hrusak and Ramos-Garcia ( have produced a model of ZFC, by iterating a Laver-type forcing, where every separable Frechet group is metrizable, and in particular, the product of Frechet groups is metrizable, and thus Frechet.

I haven't read the paper, but Justin Moore pointed me to it when I asked him about this question.

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This answers a different question since a separable sequential group need not be Frechet. – Paul Fabel Oct 19 '13 at 9:11
Can you provide an example? I can't see it right now. – Iian Smythe Oct 19 '13 at 16:50
@Iian: I mentioned this result in my answer, but it is nice that you gave the complete reference. As Paul said, a separable sequential group need not be Frechet. It is known that any Abelian group admits a sequential non-Frechet (in fact with sequential order $\omega_1$) group topology. See "Metrizability of sequential topological groups with point-countable $k$-networks" by Shibakov. – Ramiro de la Vega Oct 19 '13 at 20:07
The free group over countably many generators in the sense of Graev or Markov yields a non-Frechet, sequential, topological group. (Seek the finest topology so that x1,x2,--->1 (or x1, x2,---> the nontrivial element`infinity')) – Paul Fabel Oct 20 '13 at 0:24

A simple example of seqeuntial topological groups $G,H$ with non-seqeuntial product $G\times H$: $G=\mathbb R^\omega$ and $H=\mathbb R^\infty=\lim \mathbb R^n$ be the direct limit of finite-dimensional Euclidean spaces. The group $G$ is metrizable and $H$ is a sequential $k_\omega$-space. The non-sequentiality of $G\times H$ follows from a result [Banakh, Taras; Zdomskyĭ, Lubomyr. The topological structure of (homogeneous) spaces and groups with countable $cs^*$-character. Appl. Gen. Topol. 5 (2004), no. 1, 25--48; MR2087279] saying that each sequential topological group $G$ of countable $cs^*$-character is either metrizable or contains an open $k_\omega$-subgroup.

We say that a topological space $G$ has countable $cs^*$-character if for every point $x\in X$ there is a countable family $\mathcal N$ of subsets of $X$ such that for every neighborhood $O_x\subset X$ and every sequence $(x_n)$ convergent to $x$ in $X$ there is a set $N\subset O$ in the family $\mathcal N$ containing infinitely many points of the sequence $(x_n)$.

The mentioned result of Banakh and Zdomskyy implies that for any sequential topological group $G$ of countable $cs^*$-character the square $G\times G$ is sequential (more precisely, $G\times G$ is metrizable or contains an open $k_\omega$-subgroup).

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