Sequential space

In topology and related fields of mathematics, a sequential space is a topological space that satisfies a very weak axiom of countability. Sequential spaces are the most general class of spaces for which sequences suffice to determine the topology.

Every sequential space has countable tightness.

Definitions

Let X be a topological space and let S be a subset of X'.

• S is sequentially open if each sequence (x_n) in X converging to a point of S is eventually in S (i.e. there exists N such that x_n is in S for all n ≥ N).
• The sequential closure of S in X is the set ${\displaystyle [S]_{\text{seq}}:=\left\{x\in X:{\text{ there exists a sequence }}\left(s_{i}\right)_{i=1}^{\infty }{\text{ in }}S{\text{ such that }}\left(s_{i}\right)_{i=1}^{\infty }\to x{\text{ in }}X\right\}}$.
• S is sequentially closed if ${\displaystyle S=[S]_{\text{seq}}}$, or equivalently, if whenever (x_n) is a sequence in S converging to x, then x must also be in S.

The complement of a sequentially open set is a sequentially closed set, and vice versa. Every open subset of X is sequentially open and every closed set is sequentially closed. The converses are not generally true.

A sequential space is a space X satisfying one of the following equivalent conditions:

1. Every sequentially open subset of X is open.
2. Every sequentially closed subset of X is closed.
3. For any subset S ⊆ X that is not closed, i.e. such that ${\displaystyle {\overline {S}}\setminus S\neq \emptyset }$, there exists a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }\in S^{\mathbb {N} }}$ of elements of S that converges to an element of ${\displaystyle {\overline {S}}\setminus S}$.[1]
4. X is the quotient of a first countable space.
5. X is the quotient of a metric space.
6. For every topological space Y and every map f : X → Y, we have that f is continuous if and only if for every sequence of points (x_n) in X converging to x, we have (f(x_n)) converging to f(x).

The final equivalent condition shows that the class of sequential spaces consists precisely of those spaces whose topological structure is determined by convergent sequences in the space. Also, a space being sequential means that the original topology can be reconstructed by knowing the convergent sequences.

Proof of the equivalences —

• (${\displaystyle 1\Leftrightarrow 2}$): Assume that any sequentially open subsets is open and let F be sequentially closed. It is proved above that the complement U = X \ F is sequentially open and thus open so that F is closed. The converse is similar.

• (${\displaystyle 2\Leftrightarrow 3}$): Contraposition of 2 says that "S not closed implies S not sequentially closed", and hence there exists a sequence of elements of S that converges to a point outside of S. Since the limit is necessarily adherent to S, it is in the closure of S.

Conversely, suppose for a contradiction that a subset S := F is sequentially closed but not closed. By 3, there exists a sequence in F that converges to a point in ${\displaystyle {\overline {S}}\setminus S={\overline {F}}\setminus F}$, i.e. the limit lies outside F. This contradicts sequential closedness of F.

Sequential closure

Given a subset ${\displaystyle A\subseteq X}$ of a space ${\displaystyle X}$, the sequential closure ${\displaystyle [A]_{\text{seq}}}$ is the set

${\displaystyle [A]_{\text{seq}}=\{x\in X\mid \exists \{a_{n}\}\to x:a_{n}\in A{\text{ for all }}n\in \mathbb {N} \}}$

that is, the set of all points ${\displaystyle x\in X}$ for which there is a sequence in ${\displaystyle A}$ that converges to ${\displaystyle x}$. The map

${\displaystyle [\;]_{\text{seq}}:A\mapsto [A]_{\text{seq}}}$

is called the sequential closure operator. It shares some properties with ordinary closure, in that the empty set is sequentially closed:

${\displaystyle [\varnothing ]_{\text{seq}}=\varnothing .}$

Every closed set is sequentially closed:

${\displaystyle A\subseteq [A]_{\text{seq}}\subseteq {\overline {A}}}$

for all ${\displaystyle A\subseteq X}$; here ${\displaystyle {\overline {A}}}$ denotes the ordinary closure of the set ${\displaystyle A}$. Sequential closure commutes with union:

${\displaystyle [A\cup B]_{\text{seq}}=[A]_{\text{seq}}\cup [B]_{\text{seq}}}$

for all ${\displaystyle A,B\subseteq X}$. However, unlike ordinary closure, the sequential closure operator is not in general idempotent; that is, one may have that

${\displaystyle [A]_{\text{seq}}\subsetneq {\Big [}[A]_{\text{seq}}{\Big ]}_{\text{seq}},}$

and consequently ${\displaystyle [A]_{\text{seq}}\neq {\overline {A}}}$, even when ${\displaystyle A}$ is a subset of a sequential space ${\displaystyle X}$.

The transfinite sequential closure is defined as follows: define ${\displaystyle A_{0}}$ to be ${\displaystyle A}$, define ${\displaystyle A_{\alpha +1}}$ to be ${\displaystyle [A_{\alpha }]_{\text{seq}}}$, and for a limit ordinal ${\displaystyle \alpha }$, define ${\displaystyle A_{\alpha }}$ to be ${\displaystyle \bigcup _{\beta <\alpha }A_{\beta }}$. Then there is a smallest ordinal ${\displaystyle \alpha }$ such that ${\displaystyle A_{\alpha }=A_{\alpha +1}}$, and for this ${\displaystyle \alpha }$, ${\displaystyle A_{\alpha }}$ is called the transfinite sequential closure of ${\displaystyle A}$. (In fact, we always have ${\displaystyle \alpha \leq \omega _{1}}$, where ${\displaystyle \omega _{1}}$ is the first uncountable ordinal.) Taking the transfinite sequential closure solves the idempotency problem above.

The smallest ${\displaystyle \alpha }$ such that ${\displaystyle A_{\alpha }={\overline {A}}}$ for each ${\displaystyle A\subseteq X}$ is called sequential order of the space X.[2] This ordinal invariant is well-defined for sequential spaces.

Topology of sequentially open sets

Let ${\displaystyle \operatorname {SeqOpen} \left(X,\tau \right)}$ denote the set of all sequentially open subsets of the topological space ${\displaystyle \left(X,\tau \right)}$. Then ${\displaystyle \operatorname {SeqOpen} \left(X,\tau \right)}$ is a topology on X that contains the original topology ${\displaystyle \tau }$ (i.e. ${\displaystyle \tau \subseteq \operatorname {SeqOpen} \left(X,\tau \right)}$).

Proofs —

• Let U be sequentially open. Let us show that its complement F = X \ U is sequentially closed, i.e. that a convergent sequence (x_n)_n∈ℕ of elements of F has its limit in F.

Suppose for a contradiction that ${\displaystyle x_{n}\,{\underset {n\to \infty }{\longrightarrow }}\,x\in U}$, then there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle \left\lbrace x_{k}:k\geq N\right\rbrace \subset U}$, which contradicts the fact that all x_n are supposed to be in F.

• Conversely if F is sequentially closed, let us show that its complement U = X \ F is sequentially open.

Let (x_n)_n∈ℕ be a sequence in X such that ${\displaystyle \lim _{n\to \mathbb {N} }x_{n}=x\,\in U}$ and suppose for a contradiction that for any ${\displaystyle N\in \mathbb {N} ,\ \left\lbrace x_{k}:k\geq N\right\rbrace \not \subseteq U}$, i.e. for all ${\displaystyle N\in \mathbb {N} }$ there exists ${\displaystyle k_{N}\geq N,\ x_{k_{N}}\in F=X\setminus U}$. Define by recursion the subsequence (x_φ(n))_n∈ℕ of elements of F:  set  φ(0)= k₀  and then  φ(n+1) = k_φ(n)+1, i.e.  x_φ(0) := x_k0  and  x_φ(n+1) = x_kφ(n)+1. It is convergent as a subsequence of a convergent sequence, and all its elements are in F. Hence the limit has to be in F, which contradicts that x ∈ U. The sequence is therefore eventually in U.

• Let us show that the set of sequentially open subsets is a topology, i.e. ∅ and X are sequentially open, arbitrary unions of sequentially open subset is sequentially open and finite intersections of sequentially open subsets is sequentially open.

Any empty sequence satisfies any property and any sequence in X is eventually in X. Let (U_i)_i∈I be a family of sequentially open subsets, let ${\displaystyle U=\bigcup _{i\in I}U_{i}}$, and let (x_n)_n∈ℕ be a sequence in X converging to x ∈ U. x being in the union means there exists ${\displaystyle i_{0}\in I}$ such that ${\displaystyle x\in U_{i_{0}}}$ and by sequential openness, the sequence is eventually in U_i0. Finally, if ${\displaystyle V=\bigcap _{i=1}^{n}U_{i}}$ is a finite intersection of sequentially open subsets, then a sequence converging to x ∈ V eventually converges to each of the U_i, i.e. ${\displaystyle \ \forall \ i\in \mathbb {N} ,\ 1\leq i\leq n,\ \exists \ N_{i}\in \mathbb {N} }$ such that ${\displaystyle \left\lbrace x_{k}:k\geq N_{i}\right\rbrace \subset U_{i}}$. Taking ${\displaystyle N=\max _{1\leq i\leq n}N_{i}}$, one has ${\displaystyle \left\lbrace x_{k}:k\geq N\right\rbrace \subset V}$.

• The generated sequential topology is finer than the original one, i.e. if U is open, then it is sequentially open.

Let (x_n)_n∈ℕ be a sequence in X converging to x ∈ U. Since U is open, it is a neighborhood of x and by definition of convergence, there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle \left\lbrace x_{k},\ k\geq N\right\rbrace \subset U}$.

Properties of the topology of sequentially open sets

• The space ${\displaystyle \left(X,\operatorname {SeqOpen} \left(X,\tau \right)\right)}$ has the same convergent sequences and limits as ${\displaystyle \left(X,\tau \right)}$ (i.e. if ${\displaystyle x\in X}$ and ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ is a sequence in X, then ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }\to x}$ in ${\displaystyle \left(X,\tau \right)}$ if and only if ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }\to x}$ in ${\displaystyle \left(X,\operatorname {SeqOpen} \left(X,\tau \right)\right)}$).
• ${\displaystyle \left(X,\operatorname {SeqOpen} \left(X,\tau \right)\right)}$ is a sequential space.[3]
• If ${\displaystyle \tau _{2}}$ is any topology on X such that a sequence in X converges to a point of X in ${\displaystyle \left(X,\tau \right)}$ if and only if it does so in ${\displaystyle \left(X,\tau _{2}\right)}$, then necessarily ${\displaystyle \operatorname {SeqOpen} \left(X,\tau \right)=\operatorname {SeqOpen} \left(X,\tau _{2}\right)}$.
• If ${\displaystyle f:\left(X,\tau _{X}\right)\to \left(Y,\tau _{Y}\right)}$ is continuous, then so is ${\displaystyle f:\left(X,\operatorname {SeqOpen} \left(X,\tau _{X}\right)\right)\to \left(Y,\operatorname {SeqOpen} \left(Y,\tau _{Y}\right)\right)}$.

Fréchet–Urysohn space

Main article: Fréchet–Urysohn space

Topological spaces for which sequential closure is the same as ordinary closure are known as Fréchet–Urysohn spaces (such a space is also said to be Fréchet). That is, a Fréchet–Urysohn space has

${\displaystyle [A]_{\text{seq}}={\overline {A}}\,}$

for all ${\displaystyle A\subseteq X}$. A space is a Fréchet–Urysohn space if and only if every subspace is a sequential space. Every first-countable space is a Fréchet–Urysohn space.

The space is named after Maurice Fréchet and Pavel Urysohn.

Clearly, every Fréchet–Urysohn space is a sequential space. The opposite implication is not true in general.[4][5]

A topological space ${\displaystyle X}$ is a strong Fréchet–Urysohn space if for every point ${\displaystyle x\in X}$ and every sequence ${\displaystyle A_{1},A_{2},\ldots }$ of subsets of the space ${\displaystyle X}$ such that ${\displaystyle x\in \bigcap _{n}{\overline {A_{n}}}}$ , there are points ${\displaystyle x_{1}\in A_{1},x_{2}\in A_{2},\ldots }$ such that ${\displaystyle \{x_{n}\}\to x}$.

The above properties can be expressed as selection principles.

History

Although spaces satisfying such properties had implicitly been studied for several years, the first formal definition is originally due to S. P. Franklin (aka Stan Franklin) in 1965, who was investigating the question of "what are the classes of topological spaces that can be specified completely by the knowledge of their convergent sequences?" Franklin arrived at the definition above by noting that every first-countable space can be specified completely by the knowledge of its convergent sequences, and then he abstracted properties of first countable spaces that allowed this to be true.

Examples

Every first-countable space is sequential, hence each second-countable space, metric space, and discrete space is sequential. Every Fréchet-Urysohn space is sequential. Further examples are furnished by applying the categorical properties listed below. For example, every CW-complex is sequential, as it can be considered as a quotient of a metric space.

There are sequential spaces that are not first countable. (One example is to take the real line R and identify the set Z of integers to a point.)

An example of a space that is not sequential is the cocountable topology on an uncountable set. Every convergent sequence in such a space is eventually constant, hence every set is sequentially open. But the cocountable topology is not discrete. In fact, one could say that the cocountable topology on an uncountable set is "sequentially discrete".

The prime spectrum of a commutative Noetherian ring with the Zariski topology is sequential.

Categorical properties

The full subcategory Seq of all sequential spaces is closed under the following operations in the category Top of topological spaces:

• Quotients
• Continuous closed or open images
• Sums
• Inductive limits
• Open and closed subspaces

The category Seq is not closed under the following operations in Top:

• Continuous images
• Subspaces
• Finite products

Since they are closed under topological sums and quotients, the sequential spaces form a coreflective subcategory of the category of topological spaces. In fact, they are the coreflective hull of metrizable spaces (i.e., the smallest class of topological spaces closed under sums and quotients and containing the metrizable spaces).

The subcategory Seq is a Cartesian closed category with respect to its own product (not that of Top). The exponential objects are equipped with the (convergent sequence)-open topology. P.I. Booth and A. Tillotson have shown that Seq is the smallest Cartesian closed subcategory of Top containing the underlying topological spaces of all metric spaces, CW-complexes, and differentiable manifolds and that is closed under colimits, quotients, and other "certain reasonable identities" that Norman Steenrod described as "convenient".

See also

• Axioms of countability
• Closed graph
• First-countable space
• Fréchet–Urysohn space

References

1. Arkhangel'skii, A.V. and Pontryagin L.S.,  General Topology I, definition 9 p.12
2. • Arhangel'skiĭ, A. V.; Franklin, S. P. (1968). "Ordinal invariants for topological spaces". Michigan Math. J. 15 (3): 313–320. doi:10.1307/mmj/1029000034.
3. https://math.stackexchange.com/questions/3737020/topology-of-sequentially-open-sets-is-sequential
4. Engelking 1989, Example 1.6.18
5. Ma, Dan. "A note about the Arens' space". Retrieved 1 August 2013.

• Arkhangel'skii, A.V. and Pontryagin, L.S., General Topology I, Springer-Verlag, New York (1990) ISBN 3-540-18178-4.
• Booth, P.I. and Tillotson, A., Monoidal closed, cartesian closed and convenient categories of topological spaces Pacific J. Math., 88 (1980) pp. 35–53.
• Engelking, R., General Topology, Heldermann, Berlin (1989). Revised and completed edition.
• Franklin, S. P., "Spaces in Which Sequences Suffice", Fund. Math. 57 (1965), 107-115.
• Franklin, S. P., "Spaces in Which Sequences Suffice II", Fund. Math. 61 (1967), 51-56.
• Goreham, Anthony, "Sequential Convergence in Topological Spaces"
• Steenrod, N.E., A convenient category of topological spaces, Michigan Math. J., 14 (1967), 133-152.