TITLE: Sequential closure of a set: standard terminology, notation, and properties
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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. 
I am not a functional analyst (nor a topologist!), but it is my understanding that yes, non-sequential spaces arise in functional analysis.  Now being equipped with the necessary terminology, you should have little trouble (less than me, probably) tracking down examples.

REPLY [2 votes]: 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.