2
$\begingroup$

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

  1. $\tau_{seq}$ 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 $\tau_{seq}$-lsc. (here lsc means lower semicontinuous)

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

I would love to know more about properties of $\tau_{seq}$ 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 $\tau_{seq}$? 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. $\tau_{seq}$ 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 variation.

$\endgroup$
  • $\begingroup$ The sequential coreflexions do not interact well with the addition operation. For example, for the inductive limit $\mathbb R^\infty$ of finite-dimensional spaces and the product $\mathbb R^\infty\times\mathbb R^\omega$ the operation of addition will be discontinuous with respect to the topology $\tau_seq$. I am not sure that this is a desirable property for your purposes. $\endgroup$ – Taras Banakh Aug 23 '19 at 18:46
  • $\begingroup$ @TarasBanakh Thank you for your answer, that property is indeed could be useful to me. I must admit that I don't understand your example since I'm trained in PDE and don't have much knowledge on a general topological space. May I ask what would be the keyword to search for or a place to read more about this topology $\tau_{seq}$? I am mostly interested in the case where the original $\tau$ is the weak topology generated by bounded linear functionals on a Banach space $X$. $\endgroup$ – BigbearZzz Aug 23 '19 at 22:54
3
$\begingroup$

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.


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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It is regrettable that I couldn't find a copy of the book in our library collection but thank you very much. From reading you posts and doing some extra search I believe $w_{seq}$ is a Frechet–Urysohn space, is that correct? It seems like the following nlab page claim that it should be metrizable then. Is it true? The page doesn't give any reference to the claim. ncatlab.org/nlab/show/Frechet-Uryson+space $\endgroup$ – BigbearZzz Aug 24 '19 at 9:38
  • $\begingroup$ @BigbearZzz Now I have retyped my paper and posted it to arXiv, where it should show up in 48 hours or so. For an infinite-dimensional Banach space with separable dual, the topology $w_{seq}$ is sequential but not Frechet-Urysohn (topological groups rarely are Frechet-Urysohn). $\endgroup$ – Taras Banakh Aug 24 '19 at 9:44
  • $\begingroup$ Forgive me if this is a silly question, but from the definition of $\tau_{seq}$ shouldn't $(X,w_{seq})$ always be a sequential space? Why do we need the fact that $X^*$ is separable? ps. I am looking forward to read your paper. $\endgroup$ – BigbearZzz Aug 24 '19 at 11:27
  • $\begingroup$ Yes, $(X,w_{seq})$ is always a sequential space. The separability of $X^*$ implies the metrizability of the weak topology on bounded subsets of $X$. In this case, the topology $w_{seq}$ coincides with the weak (metrizable) topology on bounded subsets of $X$. If $X$ does not contain a copy of $\ell_1$, then the weak topology on bounded subsets of $X$ is Frechet-Urysohn (by the famous Rosenthal Theorem) and hence it coincides with the topology $w_{seq}$ on bounded sets. $\endgroup$ – Taras Banakh Aug 24 '19 at 11:40
  • $\begingroup$ On the other hand, the topology $w_{seq}$ on the Banach space $\ell_1$ coincides with the norm topology and the same is true for any Banach space with the Shur property (which means that the weak and norm convergences on the Banach space coincide). $\endgroup$ – Taras Banakh Aug 24 '19 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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