Sequential Spaces

When successive approximation is the focus

(PDF version.)

Before we talk about the topology of generalized queries, we need to return to the issue of sequential vs. topological continuity we discussed in A Digression on Topology (4): Limits and Continuity. In that post I wrote

Recall that every pseudo-metrizable space is first countable, as is every second-countable space; hence, for the sorts of spaces of interest to us, (topological) [limits]/continuity and sequential [limits]/continuity will always be the same.¹

As we’ll discuss in the next post, with generalized queries we have left the realm of first countable spaces and so we can no longer take for granted the equivalence of certain topological notions and their sequential analogs. In this post we’ll see how to ensure that the most important equivalences are retained, although some equivalences will still be lost.

Preliminaries: some terminology and notation

We’ll start by introducing or reviewing some standard terminology and notation that will facilitate the discussion.

• We write (𝑥ᵢ) ⊆ 𝑆 to mean that 𝑥ₙ ∈ 𝑆 for all 𝑛 ∈ ℕ.

• We have been taking 𝑖 implicitly to be a bound variable indicating the index of the sequence when writing (𝑥ᵢ) or even expressions such as (𝑥ₙ₊ᵢ) or (𝑓(𝑥ᵢ)); when we want to be explicit, we will indicate the bound variable as a subscript to the parenthesized expression, writing (𝑥ᵢ)ᵢ, (𝑥ₙ₊ᵢ)ᵢ, (𝑓(𝑥ᵢ))ᵢ, etc.

• We say that (𝑥ᵢ) is eventually in 𝐴 if there is some 𝑛 ∈ ℕ such that (𝑥ₙ₊ᵢ)ᵢ ⊆ 𝐴; that is, some tail of the sequence lies entirely in 𝐴.

• We may simply write 𝜏 for the topological space (𝑆, 𝜏) when discussing multiple topological spaces having the same underlying set of values 𝑆. (This echoes the usual practice of simply writing 𝑆 when the associated topology 𝜏 is clear.)

• A reminder: (𝑥ᵢ) → 𝑦 (the sequence converges to 𝑦) means that (𝑥ᵢ) is eventually in every open neighborhood of 𝑦. When discussing multiple topologies on the same set of values, we may append “in 𝜏” to disambiguate what we mean by “open set.”

• The phrase “an open 𝑈 ∋ 𝑥” means “an open neighborhood 𝑈 of 𝑥,” i.e. an open set 𝑈 such that 𝑥 ∈ 𝑈.

The results presented here are all standard, and I omit the longer, more complex proofs.

Sequentially open/closed

We start by defining sequential versions of some topological concepts:

Definition. Let 𝑆 be a topological space and 𝐴 ⊆ 𝑆.

• 𝐴 is sequentially open if for every (𝑥ᵢ) ⊆ 𝑆 and 𝑦 ∈ 𝐴, (𝑥ᵢ) → 𝑦 ⟹ (𝑥ᵢ) is eventually in 𝐴.

• 𝐴 is sequentially closed if for every (𝑥ᵢ) ⊆ 𝐴 and 𝑦 ∈ 𝑆, (𝑥ᵢ) → 𝑦 ⟹ 𝑦 ∈ 𝐴.

• The sequential closure of 𝐴 is the set of points to which some sequence in 𝐴 converges:

  \(\mathrm{scl}\, A\triangleq\left\{ y\in S\colon\exists\left(x_{i}\right)\subseteq A.\,\left(x_{i}\right)\to y\right\} .\)

  This notation leaves the topology implicit; if we want to make it explicit that “(𝑥ᵢ) → 𝑦” above means (𝑥ᵢ) → 𝑦 in 𝜏, we write scl_𝜏 𝐴.

Here are some basic properties of these concepts:

Theorem 1. For any topological space 𝑆 and 𝐴 ⊆ 𝑆,

1. 𝐴 is sequentially open ⟺ ∼𝐴 is sequentially closed;

2. 𝐴 is open ⟹ 𝐴 is sequentially open;

3. 𝐴 is closed ⟹ 𝐴 is sequentially closed;

4. 𝐴 ⊆ scl 𝐴 ⊆ cl 𝐴;

5. 𝐴 is sequentially closed ⟺ scl 𝐴 = 𝐴.

Proof. 1. ( ⇒ ) Suppose that 𝐴 is sequentially open. Then for every (𝑥ᵢ) ⊆ ∼𝐴 and 𝑦 ∈ 𝑆 such that (𝑥ᵢ) → 𝑦 we cannot have 𝑦 ∈ 𝐴, as that would imply that (𝑥ᵢ) is eventually in 𝐴, a contradiction; therefore 𝑦 ∈ ∼𝐴.

1. (⇐) Suppose that ∼𝐴 is sequentially closed but 𝐴 is not sequentially open. Then there exists (𝑥ᵢ) ⊆ 𝑆 and 𝑦 ∈ 𝐴 such that (𝑥ᵢ) → 𝑦 but 𝑥ₙ ∈ ∼𝐴 for infinitely many 𝑛. Define 𝑤ₙ to be the 𝑛-th element of (𝑥ᵢ) that belongs to ∼𝐴; then (𝑤ᵢ) is a subsequence of (𝑥ᵢ), hence (𝑤ᵢ) → 𝑦, and since (𝑤ᵢ) ⊆ ∼𝐴 this implies 𝑦 ∈ ∼𝐴, a contradiction.

2. Suppose that 𝐴 is open and (𝑥ᵢ) → 𝑦 ∈ 𝐴; then there is an open neighborhood 𝑈 of 𝑦 such that 𝑈 ⊆ 𝐴, and (𝑥ᵢ) is eventually in 𝑈, hence eventually in 𝐴.

3. Suppose that 𝐴 is closed; then ∼𝐴 is open, hence ∼𝐴 is sequentially open, hence 𝐴 is sequentially closed.

4. Let 𝑦 ∈ 𝐴. If 𝑥ₙ = 𝑦 for all 𝑛 then (𝑥ᵢ) ⊆ 𝐴 and (𝑥ᵢ) → 𝑦, so 𝑦 ∈ scl 𝐴. So 𝐴 ⊆ scl 𝐴.

Suppose 𝑦 ∈ scl 𝐴. Then there exists (𝑥ᵢ) ⊆ 𝐴 such that (𝑥ᵢ) → 𝑦. For any open 𝑈 ∋ 𝑦 we have that (𝑥ᵢ) is eventually in 𝑈, and hence 𝑈 ∩ 𝐴 ≠ 0; this implies 𝑦 ∈ 𝖼𝗅 𝐴. So scl 𝐴 ⊆ 𝖼𝗅 𝐴.

5. 𝐴 is sequentially closed iff every sequence in 𝐴 that converges, converges to a point in 𝐴. This is the same as saying that scl 𝐴 adds no new points not already in 𝐴. ∎

For the closure operator we have 𝖼𝗅 𝐴 is closed and 𝖼𝗅(𝖼𝗅 𝐴) = 𝖼𝗅 𝐴, but scl lacks the corresponding properties: scl 𝐴 is not guaranteed to be sequentially closed, nor is scl(scl 𝐴) = scl 𝐴 guaranteed to hold. The additional points added by the sequential closure may create new converging sequences that converge to points outside of the sequential closure.

If we iterate scl we can obtain an operator that behaves more like the closure operator. For any ordinal 𝛼, define

• scl⁰(𝐴) = 𝐴

• scl^{𝛼+1}(𝐴) = scl(scl^𝛼(𝐴))

• scl^𝛼(𝐴) = ⋃_{𝛽 < 𝛼} scl^𝛼(𝐴) if 𝛼 is a limit ordinal.

This transfinite sequence stabilizes by 𝛼 = 𝜔₁, the first uncountable ordinal, so

• scl^{𝜔₁}(scl^{𝜔₁}(𝐴)) = scl^{𝜔₁}(𝐴), and

• scl^{𝜔₁}(𝐴) is sequentially closed.

Review: limits and continuity

In A Digression on Topology (4): Limits and Continuity we discussed topological vs. sequential limits and continuity. Recall the following definitions, where 𝑆₁ and 𝑆₂ are topological spaces, 𝑓 : 𝑆₁ → 𝑆₂, 𝑥 ∈ 𝑆₁ and 𝑣 ∈ 𝑆₂:

• 𝑓(𝑦) → 𝑣 as 𝑦 → 𝑥 (𝑣 is a (topological) limit of 𝑓(𝑦)) if

  \(\forall\mbox{open }V\ni v.\,\exists\mbox{open }U\ni x.\,f\left[U\setminus\left\{ x\right\} \right]\subseteq V.\)

• 𝑓(𝑦) →ₛ 𝑣 as 𝑦 → 𝑥 (𝑣 is a sequential limit of 𝑓(𝑦)) if

  \(\forall\left(y_{i}\right)\subseteq S_{1}\setminus\left\{ x\right\} .\,\left(y_{i}\right)\to x\Rightarrow\left(f\left(y_{i}\right)\right)_{i}\to v.\)

• 𝑓 is (topologically) continuous at 𝑥 if 𝑓(𝑦) → 𝑓(𝑥) as 𝑦 → 𝑥. Equivalently,

  \(\forall\mbox{open }V\ni f(x).\,\exists\mbox{open }U\ni x.\,f\left[U\right]\subseteq V.\)

• 𝑓 is sequentially continuous at 𝑥 if 𝑓(𝑦) →ₛ 𝑓(𝑥) as 𝑦 → 𝑥. Equivalently,

  \(\forall\left(y_{i}\right)\subseteq S_{1}.\,\left(y_{i}\right)\to x\Rightarrow\left(f\left(y_{i}\right)\right)_{i}\to f(x).\)

• 𝑓 is (topologically) continuous if it is continuous at every point in its domain. Likewise, 𝑓 is sequentially continuous if it is sequentially continuous at every point in its domain.

Analogous to the facts that an open set is sequentially open and a closed set is sequentially closed, we have the following:

• 𝑓(𝑦) → 𝑣 as 𝑦 → 𝑥 ⟹ 𝑓(𝑦) →ₛ 𝑣 as 𝑦 → 𝑥.

• 𝑓 is continuous at 𝑥 ⟹ 𝑓 is sequentially continuous at 𝑥.

The converses do not hold in general for arbitrary topological spaces.

Definitions of “sequential space”

A sequential space is one in which continuity and sequential continuity are the same. There are, however, a variety of equivalent ways of characterizing a sequential space.

Theorem 2. The following are equivalent for any topological space 𝑆:

1. For every topological space 𝑇 and function 𝑓 : 𝑆 → 𝑇, 𝑓 is continuous ⟺ 𝑓 is sequentially continuous.

2. Every 𝐴 ⊆ 𝑆 that is sequentially open is open.

3. Every 𝐴 ⊆ 𝑆 that is sequentially closed is closed.

4. 𝑆 is a quotient of a first-countable space.

5. 𝑆 is a quotient of a metric space.

Proof. Standard results, proof omitted. ∎

When we say “𝑆 is a quotient of 𝑅” above, we mean “there is a quotient map from 𝑅 to 𝑆”. Think of the points of 𝑅 as representations of the points in 𝑆, with every point in 𝑆 having one or more representations in 𝑅. Recall that if 𝑓 : 𝑅 → 𝑆 is a quotient map then the topology of 𝑅 defines the topology of 𝑆: a set 𝑉 is open in 𝑆 iff 𝑓⁻¹[𝑉] is open in 𝑅.

Definition. A sequential space is a topological space (𝑆, 𝜏) satisfying any of the equivalent conditions 1–5 above. For an arbitrary topology 𝜏 on an arbitrary set 𝑆, we say that 𝜏 is sequential if (𝑆, 𝜏) is a sequential space.

We are most interested in property 1 above, but the other equivalent characterizations also come in handy. For instance, 2 and 3 are often easier to prove than 1; and 4 and 5 are often a more convenient characterization when dealing with computable representations.

Some properties of sequential spaces

In a sequential space, sequentially open/closed is the same as open/closed:

Theorem 3. Let 𝑆 be a sequential space and 𝐴 ⊆ 𝑆. Then

• 𝐴 is open iff 𝐴 is sequentially open;

• 𝐴 is closed iff 𝐴 is sequentially closed.

Proof. Combine items 2 and 3 of Theorem 1 with items 2 and 3 of Theorem 2.

Continuity and sequential continuity are also the same for a sequential space, according to item 1 of Theorem 2: if 𝑆 is a sequential space then for every topological space 𝑇 and function 𝑓 : 𝑆 → 𝑇, f is continuous ⟺ f is sequentially continuous . But beware! It is sequential continuity and continuity on the entire domain of the function that are equivalent, not equivalence of sequential continuity and continuity at a single point. More on this below.

Sequential topologies are entirely defined by their convergent sequences. By working only with sequential topologies we are in effect saying that what matters to us, topologically, is which sequences are convergent (and to what values):

Theorem 4. Let 𝜏 and 𝜏′ be sequential topologies on the same set 𝑆 such that (𝑥ᵢ) → 𝑦 in 𝜏 ⟺ (𝑥ᵢ) → 𝑦 in 𝜏′, for all (𝑥ᵢ) ⊆ 𝑆 and 𝑦 ∈ 𝑆. Then 𝜏 = 𝜏′.

Proof. Writing “e.i.” for “is eventually in,” a set 𝐴 ⊆ 𝑆 is sequentially open in 𝜏 iff

\(\forall\left(x_{i}\right)\subseteq S,y\in A.\,\left(x_{i}\right)\to y\mbox{ in }\tau\Rightarrow\left(x_{i}\right)\mbox{ e.i. }A.\)

But (𝑥ᵢ) → 𝑦 in 𝜏 iff (𝑥ᵢ) → 𝑦 in 𝜏′, so the above condition is equivalent to

\(\forall\left(x_{i}\right)\subseteq S,y\in A.\,\left(x_{i}\right)\to y\mbox{ in }\tau'\Rightarrow\left(x_{i}\right)\mbox{ e.i. }A,\)

which is true iff 𝐴 is sequentially open in 𝜏′. Furthermore, 𝜏 and 𝜏′ are sequential topologies, so

\(\begin{align*} & A\mbox{ is open in }\tau\\ \iff & A\mbox{ is seq. open in }\tau\\ \iff & A\mbox{ is seq. open in }\tau'\\ \iff & A\mbox{ is open in }\tau'. \end{align*}\)

∎

Being sequential is a property that is not inherited by arbitrary subspaces of a sequential space. There are, however, two important cases where sequentiality is inherited:

Theorem 5. Let 𝑆 be a sequential space and let 𝑅 ⊆ 𝑆.

1. If 𝑅 is open then 𝑅 (as a topological space) is also sequential.

2. If 𝑅 is closed then 𝑅 (as a topological space) is also sequential.

Proof. Standard result; proof given in PDF version. ∎

This will be important when dealing with the continuity of functions that are defined on only a subset of a sequential space: as long as the function’s domain is either an open or a closed subset of the sequential space, the domain forms a sequential space.

Some non-properties of sequential spaces

Now let’s consider some properties you might expect sequential spaces to have, and which first countable spaces do have, but which sequential spaces are not guaranteed to have. In the following we take 𝑆 and 𝑆′ to be sequential spaces, 𝑇 to be a topological space, 𝑓 : 𝑆 → 𝑇, 𝑥 ∈ 𝑆, and 𝑣 ∈ 𝑇.

• A subspace 𝐴 ⊆ 𝑆 is not guaranteed to be sequential, except in the special cases where 𝐴 is either open or closed.

• 𝑆 × 𝑆′ with the usual product topology is not guaranteed to be sequential.

• scl 𝐴 = 𝖼𝗅 𝐴 is not guaranteed; put another way, 𝑥 being in the closure of 𝐴 does not guarantee that there is a sequence (𝑥ᵢ) ⊆ 𝐴 converging to 𝑥. On the other hand,

  • scl 𝐴 ⊆ 𝐴 as previously mentioned;

  • (𝐴 = 𝖼𝗅 𝐴) ⟺ (𝐴 = scl 𝐴), since 𝐴 is closed iff 𝐴 = 𝖼𝗅 𝐴, and 𝐴 is sequentially closed iff 𝐴 = scl 𝐴.

• (𝑓 is continuous at 𝑥) ⟺ (𝑓 is sequentially continuous at 𝑥) is not guaranteed. On the other hand,

  • (𝑓 is continuous at 𝑥) ⟹ (𝑓 is sequentially continuous at 𝑥);

  • the global properties are equivalent: (𝑓 is continuous) ⟺ (𝑓 is sequentially continuous).

• (𝑓(𝑦) → 𝑣 as 𝑦 → 𝑥) ⟺ (𝑓(𝑦) →ₛ 𝑣 as 𝑦 → 𝑥) is not guaranteed. On the other hand,

  • (𝑓(𝑦) → 𝑣 as 𝑦 → 𝑥) ⟹ (𝑓(𝑦) →ₛ 𝑣 as 𝑦 → 𝑥) does hold.

Sequentialization

Fortunately, any non-sequential topology can be refined to a sequential topology that has exactly the same convergent sequences.

Definition. Let (𝑆, 𝜏) be a topological space. The sequentialization of (𝑆, 𝜏) is (𝑆, seq 𝜏), where

\(\mathrm{seq}\,\tau\triangleq\left\{ U\subseteq S\colon U\mbox{ is seq. open in }\left(S,\tau\right)\right\} .\)

Since all open sets are sequentially open, we immediately see that 𝜏 ⊆ seq 𝜏: we have only added additional open sets. It’s only a bit more work to verify that seq 𝜏 does in fact satisfy the requirements of a topology, and that it is in fact sequential. We start with a simple lemma.

Lemma 6. If (𝑥ᵢ) → 𝑦 in 𝜏 then (𝑥ᵢ) → 𝑦 in seq 𝜏.

Proof. Assume that (𝑥ᵢ) → 𝑦 in 𝜏. Let 𝑈 be any open neighborhood of 𝑦 in seq 𝜏; that is, 𝑦 ∈ 𝑈 and 𝑈 is sequentially open in 𝜏. (𝑥ᵢ) is eventually in 𝑈, by definition of “sequentially open.” This holds for any open neighborhood of 𝑦 in seq 𝜏, therefore (𝑥ᵢ) → 𝑦 in seq 𝜏.

Theorem 7. If 𝜏 is a topology on 𝑆 then seq 𝜏 is a sequential topology on 𝑆.

Proof. Let’s check that seq 𝜏 is in fact a topology:

1. ∅ ∈ seq 𝜏 and 𝑆 ∈ seq 𝜏: these sets are in 𝜏, hence in seq 𝜏.

2. Let 𝒰 be any subset of seq 𝜏 and let 𝑈 = ⋃𝒰 be the union of the sets in 𝒰. If (𝑥ᵢ) → 𝑦 in 𝜏 and 𝑦 ∈ 𝑈, then 𝑦 ∈ 𝑉 for some 𝑉 ∈ 𝒰, which is sequentially open in 𝜏, hence (𝑥ᵢ) is eventually in 𝑉 ⊆ 𝑈. Thus 𝑈 is sequentially open in 𝜏 and therefore belongs to seq 𝜏.

3. Let 𝑈, 𝑉 ∈ seq 𝜏. If (𝑥ᵢ) → 𝑦 in 𝜏 and 𝑦 ∈ 𝑈 ∩ 𝑉, then since 𝑈 and 𝑉 are both sequentially open in 𝜏, there exist 𝑚 and 𝑛 such that 𝑥ₖ ∈ 𝑈 for 𝑘 ≥ 𝑚 and 𝑥ₖ ∈ 𝑉 for 𝑘 ≥ 𝑛, hence 𝑥ₖ ∈ 𝑈 ∩ 𝑉 for 𝑘 ≥ max(𝑚,𝑛). So 𝑈 ∩ 𝑉 is sequentially open in 𝜏 and therefore belongs to seq 𝜏.

To show that seq 𝜏 is in fact sequential, we have to check that every set that is sequentially open in seq 𝜏 is open in seq 𝜏, i.e., sequentially open in 𝜏. So suppose 𝐴 is sequentially open in seq 𝜏. Let (𝑥ᵢ) → 𝑦 ∈ 𝐴 in 𝜏; then by Lemma 6, (𝑥ᵢ) → 𝑦 in seq 𝜏, and hence (𝑥ᵢ) is eventually in 𝐴 (which is sequentially open in seq 𝜏). This holds for arbitrary (𝑥ᵢ) → 𝑦 ∈ 𝐴 in 𝜏, so 𝐴 is sequentially open in 𝜏. ∎

An important property of seq 𝜏 is that it has exactly the same converging sequences as 𝜏:

Theorem 8. Let (𝑆, 𝜏) be a topological space. Then (𝑥ᵢ) → 𝑦 in 𝜏 ⟺ (𝑥ᵢ) → 𝑦 in seq 𝜏.

Proof. (⇐) Follows directly from 𝜏 ⊆ seq 𝜏. (⇒) See Lemma 6. ∎

Another important property of seq is that it leaves sequential topologies unchanged; thus if you’re not sure whether 𝜏 is sequential, and you only want to expand it if this is necessary to make it sequential, it is always safe to take seq 𝜏:

Theorem 9. Let (𝑆, 𝜏) be a sequential space; then seq 𝜏 = 𝜏.

Proof. For any 𝐴 ⊆ 𝑆,

\(\begin{align*} & A\mbox{ is open in }\mathrm{seq}\,\tau\\ \iff & A\mbox{ is seq. open in }\tau\\ \iff & A\mbox{ is open in }\tau. \end{align*}\)

∎

Finally, we’ll note that seq may be defined in terms of the transfinitely iterated sequential closure scl^{𝜔₁}:

Theorem 10. Let (𝑆, 𝜏) be a topological space and define 𝜏′ to be the unique topology on 𝑆 whose closed sets are exactly the fixed points of scl_𝜏^{𝜔₁}:

\(A\mbox{ is closed in }\tau'\iff\mathrm{scl}_{\tau}^{\omega_{1}}\left(A\right)=A.\)

𝜏′ is well defined (there is exactly one topology on 𝑆 satisfying this condition) and 𝜏′ = seq 𝜏.

Proof. Too involved to give here. It involves the notion of a Kuratowski closure operator, the properties of such operators, and showing that scl_𝜏^{𝜔₁} is a Kuratowski closure operator. ∎

Product spaces and multi-argument functions

When investigating the joint continuity of all arguments to a multi-argument function, such as the AND (∧) of two generalized queries, we are dealing with the product of the spaces for each argument. But, as mentioned above, even though 𝑆₁,…,𝑆ₙ may all be sequential spaces, this does not guarantee that the product space²

\(S_1\times\cdots\times S_n\)

is sequential.

Sequentialization comes to the rescue here: rather than assume the standard product space topology, we work with the sequential space

\(\mathrm{seq}\left(S_{1}\times\cdots\times S_{n}\right).\)

Sequentialization and product play together nicely in certain ways:

• The sequential product is associative, up to homeomorphism:

  \(\begin{align*} &\mathrm{seq}\left(S_{1}\times\mathrm{seq}\left(S_{2}\times S_{3}\right)\right) \\ {} \cong {}& \mathrm{seq}\left(\mathrm{seq}\left(S_{1}\times S_{2}\right)\times S_{3}\right), \end{align*}\)

  meaning that, even though the two topological spaces are not strictly the same, they are homeomorphic (via the map ((𝑥,𝑦),𝑧) ↦ (𝑥,(𝑦,𝑧)).) This echoes the same property of the topological product.

• The sequential product yields the same result whether or not you sequentialize any of its arguments:

  \(\begin{align*} \mathrm{seq}\left(S_{1}\times S_{2}\right) & =\mathrm{seq}\left(\mathrm{seq}\, S_{1}\times S_{2}\right)\\ & =\mathrm{seq}\left(\mathrm{seq}\, S_{1}\times\mathrm{seq}\, S_{2}\right)\\ & =\mathrm{seq}\left(S_{1}\times\mathrm{seq}\, S_{2}\right). \end{align*}\)

• All of this generalizes to 𝑛-fold products seq(𝑆₁ × ⋯ × 𝑆ₙ) for 𝑛 > 2.

Upcoming

Now that we have in hand the notion of a sequential space and the sequentialization of a topology, in the next post we can define the topology of generalized premises. It will amount to the sequential topology in which the sequence of generalized queries (ℰᵢ) converges iff the sequence of sets of truth assignments ([ℰᵢ]) converges pointwise.

1. There was an error of terminology in the original statement, which I have fixed here.

2. Recall that the topology of the product space has as a base all sets of form 𝑈₁ × ⋯ × 𝑈ₙ, where each 𝑈ᵢ is open in 𝑆ᵢ.