# Sequential closure of a set: standard terminology, notation, and properties

Let $X$ be a topological vector space (or, perhaps, more generally uniform space). Let $A\subset X$ be a subset. Let $A^s$ denote the set of limits of all convergent sequences (I guess $A^s$ is called sequential closure of $A$, but I am not sure). If $X$ is not metrizable, then $A^s$ may not be closed or even sequentially closed. Thus one may repeat the procedure of taking "sequential closure".

By transfinite induction, for any ordinal $\omega$ one defines $A^\omega$ as follows:

1) if $\omega=1$ then $A^\omega:=A^s$;

2) if $\omega=\omega'+1$ then $A^\omega=(A^{\omega'})^s$;

3) if $\omega$ is a limit ordinal then $A^\omega=\cup_{\omega'<\omega}A^{\omega'}$.

It is easy to see that there exists sufficiently large ordinal $\eta_0$ such that the process stabilizes: for any $\eta\geq \eta_0$ one has $A^{\eta}=A^{\eta_0}$. In particular $A^{\eta_0}$ is sequentially closed.

QUESTIONS. (1) Is there a standard name and notation of $A^{\eta_0}$?

(2) Does $A^{\eta_0}$ have to be topologically closed? Particularly I am interested in the situation when $X$ is a locally convex vector space, and $A$ is a linear subspace.

• You probably already know of it, but in case not I mention that the paper arxiv.org/abs/math/0203139 of Ostrovskii and the references contained therein may be of interest; consideration of this idea was included in the appendix to Banach's book. Jun 8, 2014 at 6:50
• @PhilipBrooker: thanks for the reference. As I learned from it just now (and already checked in the source), the sequential closures of transfinite order (this is the terminology used in the paper) were already introduced by Banach in his classical book " Th/'eorie des op/'erations lineares", 1932 (see p. 213).
– asv
Jun 8, 2014 at 13:54

This answer on math.SE discusses the operator $$A \mapsto A^{\eta_0}$$ obtained by transfinitely iterating the sequential closure until stabilization is reached. It gives it the notation $$\operatorname{scl}^*$$. I don't know if the notation is standard, but the author of that answer is a general topologist, so there's some authority there.
By the way, the least ordinal $$\eta_0$$ such that iterating the sequential closure $$\eta_0$$ times gives you $$\operatorname{scl}^*$$ is called the sequential order of the space.
You ask if $$\operatorname{scl}^*(A)$$ is necessarily closed. Since $$A \subset \operatorname{scl}^*(A) \subset \overline{A}$$, this happens iff $$\operatorname{scl}^*(A) = \overline{A}$$: a space satisfying this property for all subsets is called sequential. These facts, together with some references, appear in $$\S$$ 2.2 of my notes on convergence. Perhaps the easiest example (= Example 2.1.5 in my notes) of a nonsequential space is an uncountable set endowed with the cocountable topology, in which a proper subset is closed precisely when it is countable. This space is sequentially discrete -- the only convergent sequences are the eventually constant ones -- so every subset is sequentially closed.
Let $I$ be an uncountable set, and give $\mathbb{R}^{I}$ the product topology. Then $\mathbb{R}^{I}$ becomes a complete locally convex topological vector space. Let $A\subseteq\mathbb{R}^{I}$ be the set of all sequences $(x_{i})_{i\in I}$ such that $x_{i}\neq 0$ for at most countably many indices. Then $A$ is a sequentially closed, but $A$ is dense in $\mathbb{R}^{I}$, so $A$ is not closed.