A Digression on Topology (2)

Constructing and describing topologies

In Part 1 we discussed the definition of a topology and the basics of open sets and closed sets. Now let’s look at common ways of constructing or describing topologies.

Subspaces

Sometimes we want to restrict a topological space to a subset, while keeping the topology compatible with the original.

Definition. If (𝑆, 𝜏) is a topological space and 𝑅 ⊆ 𝑆, then the subspace topology on R is defined as

\(\tau'\triangleq\left\{ O\cap R\colon O\in\tau\right\} ,\)

and (𝑅, 𝜏′) is called a subspace of (𝑆, 𝜏). We sometimes call 𝑂 ∩ 𝑅 the restriction of 𝑂 to 𝑅.

It is straightforward to show that the subspace topology as defined above does in fact satisfy the requirements of a topology on 𝑅.

It is customary when talking about a subset of a topological space to assume that the subspace topology is intended, unless otherwise specified.

Example. The rational numbers ℚ are a subset of the real numbers ℝ. The subspace topology on ℚ induced by the standard topology on ℝ has, as its open sets, all sets obtained by taking an open set in ℝ and removing all the irrational numbers. So, among others, all sets of the form

\(\left(a,b\right)\cap\mathbb{Q}=\left\{ x\in\mathbb{Q}\colon a<x<b\right\} \)

for some 𝑎,𝑏 ∈ ℝ are open sets in this subspace topology.

Example. An open set in the subspace topology need not be open in the topology of the original space. For the subspace topology on the nonnegative reals [0, ∞), any half-open interval of the form [0, 𝑏) where 𝑏 ≥ 0, is an open set, since

\(\left[0,b\right)=\left(-1,b\right)\cap\left[0,\infty\right).\)

Bases

The next method of construction we have already used in defining the topology induced by a pseudo-metric.

Definition. A base of a topology 𝜏 is a collection of open sets ℬ such that every open set in 𝜏 is the union of some sub-collection of ℬ:

\(\tau=\left\{ \bigcup\mathcal{C}\colon\mathcal{C}\subseteq\mathcal{B}\right\} . \)

In this case we say that ℬ generates 𝜏.

Note that we say a base, not the base, as a topology will have many possible bases. The collection of open balls

\(\mathcal{B}_{0}\triangleq\left\{ B_{d}\left(x,r\right)\colon x\in S,r>0\right\} \)

is a base for the topology induced by a pseudo-metric 𝑑 on a set 𝑆. But this smaller collection is also a base:

\(\mathcal{B}_{1}\triangleq\left\{ B_{d}\left(x,r\right)\colon x\in S,r>0,r\in\mathbb{Q}\right\} ,\)

that is, the collection of those open balls whose radius is a rational number. For the standard topology on ℝⁿ, the following even smaller, countable collection is a base:

\(\mathcal{B}_{2}\triangleq\left\{ B_{e}\left(x,r\right)\colon x\in\mathbb{Q}^{n},r\in\mathbb{Q}\right\} ,\)

that is, the collection of those open balls for which both the radius and the coordinates of the center point are rational. The collection of all open boxes is also a base of the standard topology on ℝ:

\(\begin{align*} \mathcal{B}_{3} & \triangleq\left\{ \beta\left(\boldsymbol{a},\boldsymbol{b}\right)\colon \boldsymbol{a},\boldsymbol{b}\in\mathbb{Q}^{n}\right\} \\ \beta\left(\boldsymbol{a},\boldsymbol{b}\right) & \triangleq\left\{ \boldsymbol{x}\in\mathbb{R}^{n}\colon a_{i}<x_{i}<b_{i},1\leq i\leq n\right\} . \end{align*}\)

Not every collection of sets ℬ generates a topology; the result of taking all unions of subcollections of ℬ must satisfy the three properties defining a topology. The following theorem gives us necessary and sufficient criteria for a collection of subsets of a set 𝑆 to generate a topology on 𝑆.

Theorem. If ℬ is a base for a topology (𝑆, 𝜏), then

1. the elements of ℬ cover 𝑆: every 𝑥 ∈ 𝑆 belongs to some 𝐵 ∈ ℬ; and

2. for every 𝐴,𝐵 ∈ ℬ and 𝑥 ∈ 𝐴 ∩ 𝐵, there exists some 𝐶 ∈ ℬ such that 𝑥 ∈ 𝐶 ⊆ 𝐴 ∩ 𝐵.

Conversely, if ℬ is a collection of subsets of 𝑆 satisfying (1) and (2), then 𝜏, the set of all unions of subcollections of ℬ, is a topology on 𝑆 (called the topology generated by ℬ), and ℬ is a base for 𝜏.

The intuition behind (2) is that 𝐴 ∩ 𝐵 must be an open set, and hence must be expressible as the union of sets from ℬ. Using this theorem we can show that

Theorem. The topology induced by a pseudo-metric 𝑑 on a set 𝑆 is, in fact, a topology on 𝑆.

Proof. Clearly the open balls cover 𝑆, since 𝑥 ∈ 𝐵(𝑥, 𝑟) for any 𝑥 and 𝑟 > 0. For (2) we must show that, given open balls 𝐴 = 𝐵(𝑎, 𝑟) and 𝐵 = 𝐵(𝑏, 𝑠), and point 𝑥 ∈ 𝐴 ∩ 𝐵, the point 𝑥 belongs to an open ball contained entirely within 𝐴 ∩ 𝐵.

We necessarily have 𝑑(𝑎, 𝑥) < 𝑟 and 𝑑(𝑏, 𝑥) < 𝑠. Defining

\(t\triangleq\min\left(r-d\left(a,x\right),s-d\left(b,x\right)\right)\)

we have by the triangle inequality that every point in 𝐵(𝑥, 𝑡) is within distance 𝑟 of 𝑎 and within distance 𝑠 of 𝑏, i.e., 𝐵(𝑥, 𝑡) ⊆ 𝐴 ∩ 𝐵. ∎

Subbases

Sometimes even a base for a topology gets complicated to describe, and it helps to be construct it from simpler elements: a sub-base.

Definition. A subbase of a topology 𝜏 on a set 𝑆 is a collection of open sets 𝔅 ⊆ 𝜏 satisfying either one of the following two equivalent conditions:

• 𝜏 is the smallest topology containing 𝔅; or

• the collection of all finite intersections of elements of 𝔅 is a base for 𝜏.

In the above definition we use the convention that ⋂∅ = 𝑆. The reasoning is as follows. The elements of 𝔅 are all subsets of 𝑆, and for any set 𝐵 ⊆ 𝑆 we have 𝑆 ∩ 𝐵 = 𝐵. Then for 𝐴,𝐵,𝐶 ⊆ 𝑆,

\(\begin{align*} \bigcap\left\{ A,B,C\right\} & =S\cap A\cap B\cap C\\ \bigcap\left\{ A,B\right\} & =S\cap A\cap B\\ \bigcap\left\{ A\right\} & =S\cap A\\ \bigcap\left\{ \right\} & =S. \end{align*}\)

Note that any base ℬ for a topology 𝜏 is also a subbase for 𝜏: the collection of all finite intersections of elements of ℬ includes ℬ itself, as ⋃{𝐴} = 𝐴, and if you add additional open sets to a base it is still a base.

We earlier mentioned that a point in a topological space can be described in terms of the open sets to which it belongs; if two points belong to the same open sets, they’re topologically equivalent / indistinguishable. If we have a subbase for the topology, we can restrict our attention to just the members of the subbase:

Theorem. Let 𝔅 be a subbase for a topological space. Two points 𝑥 and 𝑦 in the space are topologically equivalent iff they belong to the same subbasic open sets: for all 𝐵 ∈ 𝔅, both 𝑥,𝑦 ∈ 𝐵 or both 𝑥,𝑦 ∉ 𝐵.

Because of this the members of a subbase are sometimes called atomic properties. If a topological space is generated by a countable subbase 𝔅 = { 𝐵₀, 𝐵₁, 𝐵₂, … } , then for any point 𝑥 its equivalence class [𝑥] is identified by a sequence of bits specifying the properties it does and does not have: (𝑏₀, 𝑏₁, 𝑏₂, …) where 𝑏ᵢ = 1 if 𝑥 ∈ 𝐵ᵢ and 𝑏ᵢ = 0 if 𝑥 ∉ 𝐵ᵢ.

Example. If we define 𝔅 to be the collection of all open sets of the form (𝑎, ∞) or (-∞, 𝑎), then 𝔅 is a subbase of the standard topology on ℝ induced by the Euclidean metric: any open interval (𝑎, 𝑏) can be expressed as (𝑎, ∞) ∩ (-∞, 𝑏), and the open intervals form a base for the standard topology on ℝ. In fact, if we restrict our attention to just those elements of 𝔅 for which 𝑎 is rational, this smaller collection is also a subbase, one that is countable. This gives us a countable set of atomic properties of the form 𝑥 < 𝑎 or 𝑥 > 𝑎 for some 𝑎 ∈ ℚ, and any real number 𝑥 is identified by those atomic properties it satisfies.

Example. Let 𝔅 be the collection of all half-spaces of ℝ² orthogonal to one of the axes. These are sets of one the following four forms:

• all points (𝑥, 𝑦) with 𝑥 > 𝑎;

• all points (𝑥, 𝑦) with 𝑥 < 𝑎;

• all points (𝑥, 𝑦) with 𝑦 > 𝑎; or

• all points (𝑥, 𝑦) with 𝑦 < 𝑎.

The set of all finite intersections of members of 𝔅 is just the collection of open boxes 𝛽(𝒂,𝒃), where 𝒂 ∈ (ℝ ∪ { -∞} )² and 𝒃 ∈ (ℝ ∪ { ∞} )². As previously mentioned, the open boxes form a base for the standard topology on ℝ², so 𝔅 is a subbase.

We saw previously that we could describe a topology 𝜏 on a set 𝑆 by specifying a base ℬ for 𝜏, but we had to verify that ℬ satisfied certain conditions. We can also describe a topology 𝜏 by specifying a subbase 𝔅 for 𝜏; what conditions must 𝔅 satisfy?

Conveniently, any collection of subsets of 𝑆 will do.

Theorem. Suppose that 𝑆 is a set and 𝔅 is any collection of subsets of 𝑆. Then 𝔅 is a subbase of a topology 𝜏 on 𝑆. That is, if we define ℬ to be the collection of all finite intersections of elements of 𝔅 and define 𝜏 to be the collection of all unions of elements of ℬ, then 𝜏 is a topology on 𝑆 and ℬ is a base for 𝜏.

Product spaces

If we have topologies for sets 𝑆₁ and 𝑆₂, they define an obvious topology on the Cartesian product 𝑆₁ × 𝑆₂ (the set of all ordered pairs (𝑥, 𝑦) with 𝑥 ∈ 𝑆₁ and 𝑦 ∈ 𝑆₂.)

Definition. Let (𝑆₁, 𝜏₁) and (𝑆₂, 𝜏₂) be topological spaces. Let

\(\mathcal{B}\triangleq\left\{ U_{1}\times U_{2}\colon U_{1}\in\tau_{1},U_{2}\in\tau_{2}\right\} ;\)

that is, the sets in ℬ are formed by taking one open set from each component space and forming their Cartesian product. Then the product topology on the set 𝑆₁ × 𝑆₂ is the topology for which ℬ is a base.

For 𝑛 > 2 topological spaces (𝑆ᵢ, 𝜏ᵢ) we define their product topology on the set 𝑆₁ × ⋯ × 𝑆ₙ analogously.

Note that if we define 𝔅 to be the collection of all sets of the form 𝑈₁ × 𝑆₂ or 𝑆₁ × 𝑈₂ for 𝑈₁ ∈ 𝜏₁ and 𝑈₂ ∈ 𝜏₂, then any finite intersection of elements of 𝔅 has the form

\(\left(U_{1}\cap\cdots\cap U_{m}\right)\times\left(V_{1}\cap\cdots \cap V_{n}\right)\)

where 𝑚, 𝑛 ≥ 0, 𝑈ᵢ ∈ 𝜏₁ for 1 ≤ 𝑖 ≤ 𝑚, and 𝑉ᵢ ∈ 𝜏₂ for 1 ≤ 𝑖 ≤ 𝑛; but since the product of open sets is also an open set, this means that the collection of all finite intersections of elements of 𝔅 is just the collection ℬ. Thus 𝔅 is a subbase for the product topology on 𝑆₁ × 𝑆₂, which verifies that it is in fact a topology.