Properties of the topology of sequential convergence $\tau_\text{seq}$

Question

Let $(X,\tau)$ be a Hausdorff space. Denote by $\tau_\text{seq}$ the topology on $X$ whose closed sets are the sequentially $\tau$-closed subsets of $X$. I have read that $\tau_\text{seq}$ has the following properties:

1. $\tau_\text{seq}$ is the strongest topology on $X$ for which the converging sequences are the $\tau$-converging sequences.

2. $f:X\to\Bbb R$ is sequentially $\tau$-lsc $\iff$ $f$ is $\tau_\text{seq}$-lsc. (Here lsc means lower semicontinuous.)

3. $\tau_\text{seq}=\tau$ if $\tau$ is a first-countable topology.

I would love to know more about properties of $\tau_\text{seq}$ but the reference for the above claims is the article Topologie e strutture di convergenza by Dolcher and, unfortunately, I don't read or speak Italian.

Does anyone know a book/paper/article written in English that discusses the properties of $\tau_\text{seq}$? For example, whether it is first-countable, completely regular, or locally convex, provided that we know what $\tau $ is.

The particular case that is of interest to me is when $X$ is an infinite dimensional Banach space and $\tau = w$, the weak topology. $\tau_\text{seq}$ is very related to the sequential $w$-lsc relaxation of a (nonlinear) functional $J:X\to \Bbb R$. This kind of relaxation is, in general, different from the topological $w$-lsc relaxation and is more useful in optimization and calculus of variations.

Answer 1

Concerning the sequential coreflexion $w_{seq}$ of the weak topology on a Banach space $X$ the following characterization can be proved.

Theorem. For a Banach space $X$ the following conditions are equivalent:

1) $X$ is reflexive;

2) $(X,w_{seq})$ is a locally convex topological vector space;

3) the addition operation $+:X\times X\to X$ is jointly continuous with respect to the topology $w_{seq}$.

Proof. (1)$\Rightarrow$(2) If $X$ is reflexive, then the closed unit ball $B$ of $X$ is compact in the weak topology. Moreover, it is Eberlein compact and hence Frechet-Urysohn, which implies that on each ball $n\cdot B$ the topology $w_{seq}$ induces the weak topology. Since each weakly convergent sequence is bounded, the topology $w_{seq}$ coincides with the topology of the direct limit $\varinjlim nB$ of the sequence $(n B)_{n\in\mathbb N}$, which implies that $(X,w_{seq})$ is a $k_\omega$-space. Now the continuity of the addition $+:nB\times nB\to 2nB$ in the weak topology implies the continuity of the addition in the topology $w_{seq}$ of direct limit $\varinjlim n\cdot B=(X,w_{seq})$. By the same reason, the multiplication map $X\times\mathbb R\to X$ is continuous with respect to the topology $w_{seq}$. So, $(X,w_{seq})$ is a linear topological space. Its local convexity can be proved using the local convexity of the weak topology and the coincidence of the topology $w_{seq}$ with the direct limit topology $\varinjlim nB$ of the sequence of compact convex sets.

(2)$\Rightarrow$(3) is trivial.

(3)$\Rightarrow$(1) Assumining that $X$ is not reflexive, we conclude that the closed unit ball endowed with the weak topology is not compact and hence not sequentially compact (by the classical Eberlian-Smulian Theorem). Consequently, $X$ contains a non-reflexive separable Banach subspace $Y$. Assuming that the addition operation is continuous with respect to the topology $w_{seq}$ on $X$, we conclude that it is continuous with respect to the topology $w_{seq}$ on $Y$ and $(Y,w_{seq})$ is a topological group.

The separability of $Y$ implies that the closed ball $nB_Y$ of radius $n$ is metrizable and separable in the topology $w_{seq}$ which coincides with the weak topology on $nB$. Consequently, the metrizable separable space $(nB,w_{seq})$ has countable base $\mathcal B_n$ of the (weak) topology. The union $\mathcal B=\bigcup_{n\in\mathbb N}\mathcal B_n$ is a countable $cs$-network at zero of the space $(Y,w_{seq})$. The latter means that for any sequence $\{x_n\}_{n\in\omega}\subset X$ that converges to zero in the topology $w_{seq}$ and any neighborhood $U\in w_{seq}$ of zero there exists a set $B\in\mathcal B_n$ such that $0\in B\subset U$ and $B$ contains all but finitely many points $x_n$.

By a result of Banakh and Zdomskyy, a sequential topological group having a countable $cs$-network at zero is either metrizable of contains an open $k_\omega$-subgroup. But $(Y,w_{seq})$ is neighter metrizable nor contains an open $k_\omega$-subgroup. This contradiction shows that $(Y,w_{seq})$ is not a topological group and the addition is discontinuous.

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Remark. Some topologies near to $w_{seq}$ have been considered in the paper [T.Banakh, On topological classification of normed spaces endowed with the weak topology or the topology of compact convergence], published in this book.

Answer 2

I am aware of two general topology textbooks which discuss sequential spaces: (probably there is also discussion in earlier works by e.g. Frechet, Kuratowski...)

1. General Topology I by Arkhangel'skii and Fedorchuck;
2. General Topology by Engelking.

The first contains a few basic results with proofs; as I recall, the second mostly leaves things as an exercise for the reader.

I confess this is a somewhat low-value answer, in that I found both of these books by looking at the references in the Wikipedia page regarding sequential spaces!

(Also, my impression, from attempting to read the article by Dolcher with what Italian I know, is that the approach there is very abstract, and that you are probably better off just trying to prove things yourself, or asking someone like Taras Banakh for help. :) )