4
$\begingroup$

Consider the usual sequential modifications of topologies (spaces) in the categories of topological spaces $\text{Top}$, topological vector spaces $\text{TVS}$ and locally convex spaces $\text{LCS}$ :

Let $X$ be a real vector space. All of the occurring topologies are assumed to be Hausdorff. If $\tau$ is a topology, linear topology, locally convex topology on $X$ then denote by $\tau_s$, $\tau_l$ and $\tau_{lc}$ the respectively finest topology, finest linear topology and finest locally convex topology with the same convergent sequences as $\tau$. Then $\tau \subseteq \tau_s$, if $\tau$ is linear then $\tau \subseteq \tau_l \subseteq \tau_s$ and if $\tau$ is locally convex then $\tau \subseteq \tau_{lc} \subseteq \tau_l \subseteq \tau_s$. In general, $\tau_s$ needs not be a locally convex and not even a linear topology. These modifications of topologies lead to coreflective subcategories of the above mentioned categories $\text{Top}$, $\text{TVS}$ and $\text{LCS}$.

If $(X, \tau)$ is a topological space then $(X, \tau_s)$ is a sequential space. In general, a finite product of sequential spaces needs not be sequential. (Even worse: the product of a Fréchet-Urysohn space with a compact metric space needs not be sequential (see Example 1).)

If $(X, \tau)$ is a locally convex space then $(X, \tau_{lc})$ is a convex-sequential ($C$-sequential) space. Such spaces were first investigated by [Dudley, "On sequential convergence" (1964)]. Dudley (Theorem 6.2) shows that convex-sequential spaces are finitely productive. More generally, [Hus̆ek, "Mazur-like topological linear spaces and their products" (1997)] shows that a $\kappa$-fold product of convex-sequential spaces is convex-sequential if and only if $\kappa$ is a nonmeasurable cardinal. In particular, countable or continuum products of convex-sequential spaces are convex-sequential. This result (and many similar results for such Mazur-like spaces) can be seen as a generalization of the Mackey-Ulam theorem (for bornological spaces).

If $(X, \tau)$ is a topological vector space then let us call the space $(X, \tau_l)$ a linear-sequential space. Such spaces were investigated by [Katsaras, Benekas, "Sequential convergence in topological vector spaces" (1995)] and by [Ferrer, Morales, Sanchez Ruiz, "Sequential convergence in topological vector spaces" (2000)] (yes, same title!). Both papers are independent but have many similar results. [Katsaras, Benekas] refer to such linear-sequential spaces simply as "sequential" and show that these are finitely productive (since these spaces are permanent under inductive limits and thus under direct sums). [Ferrer, Morales, Sanchez Ruiz] refer to such spaces linear-sequential spaces as "sequentially maximal" and show more generally that they are productive, i.e. an arbitrary product of linear-sequential spaces is linear-sequential.

Question: What is the main reason that in the world of topological vector spaces the linear-sequentialness property is permanent under arbitrary products, whereas in the (basically simpler) locally convex world the permanence of the convex-sequentialness property under products depends on the size of the factors?

[Hus̆ek] also briefly mentions that the non-locally convex case is different from the locally convex case and has a follow-up paper [Hus̆ek, "Productivity of some classes of topological linear spaces" (1997)] that deals with the non-locally convex case. But I'm not a set theorist and don't really grasp the main difference. Also suprisingly, in this paper, Hus̆ek mentions in his abstract (see also Corollary 9) that "topological linear spaces closed under quotients and inductive limits are either not countably productive or nonmeasurably productive."

So, is there is some "contradiction" to the result of [Ferrer, Morales, Sanchez-Ruiz]?

$\endgroup$

1 Answer 1

2
$\begingroup$

I think the result of [Ferrer, Morales, Sanchez-Ruiz] is simply wrong. Mazur announced in

Sur la structure des fonctionelles linéaires dans certains espaces (L), Ann. Soc. Polon. Math. 19 (1946), 241

that every sequentially continuous linear functional on $\mathbb{R}^I$ is continuous iff $|I|$ is not a (two-value) measurable cardinal, but the original proof was never published.

A slight generalization of the argument for the direction that disproves [Ferrer, Morales, Sanchez-Ruiz] is found in the proof of Theorem 2 of another paper of Hušek. He proves the stronger statement for linear functionals that are continuous on well-ordered sequences of length less than a measurable cardinal rather than merely continuous on sequences, but the argument isn't any different than what you would use for sequences alone.

Since $\mathbb{R}^I$ can be viewed as $C(I, \mathbb{R})$ where $I$ is given the discrete topology and $C(I, \mathbb{R})$ is given the topology of pointwise convergence, it is natural to characterize the completely regular spaces $X$ for which every sequentially continuous linear functional on $C(X, \mathbb{R})$ is continuous. It turns out that this occurs precisely when $X$ is realcompact, which is equivalent to it being embeddable into some power of $\mathbb{R}$. It is independently known that a discrete space $I$ is realcompact iff $|I|$ is nonmeasurable.

This theorem on $C(X, \mathbb{R})$ is semifolklore and was first published by Mrowka, but a proof appears in these notes of Wilansky, including a characterization of the sequentially continuous linear functionals on $C(X, \mathbb{R})$ as linear combinations of evaluations at points of the realcompactification of $X$.

$\endgroup$
1
  • $\begingroup$ Indeed, there is an error in the proof of [Ferrer, Morales, Sanchez Ruiz, "Sequential convergence in topological vector spaces" (2000), Proposition 3.2]. However, their proof works for countable products and it follows then by [Hus̆ek, "Productivity of some classes of topological linear spaces" (1997), Theorems 2 and 3] that this category is then already productive up to the first sequential cardinal (a large cardinal) but not more. $\endgroup$
    – yada
    Nov 25, 2016 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.