§ 13. Basis for a Topology

Definition 13.1 : Basis / Topology Generated by Basis

If \(X\) is a set, a basis for a topology on \(X\) is a collection \(\mathcal{B}\) of subsets of \(X\) (called basis elements) such that

1. For each \(x \in X\), there is at least one basis element \(B\) containing \(x\).
2. If \(x\) belongs to the intersection of two basis elements \(B_{1}\) and \(B_{2}\), then there is a basis element \(B_{3}\) containing \(x\) such that \(B_{3} \subset B_{1} \cap B_{2}\).

(Basis \(\Rightarrow\) Topology)
If \(\mathcal{B}\) satisfies these two conditions, then we define the topology \(\mathcal{T}\) generated by \(\mathcal{B}\) as follows:
A subset \(U\) of \(X\) is said to be open in \(X\) (that is, to be an element of \(\mathcal{T}\) ) if for each \(x \in U\), there is a basis element \(B \in \mathcal{B}\) such that \(x \in B\) and \(B \subset U\). Note that each basis element is itself an element of \(\mathcal{T}\).

Example

1. Let \(\mathcal{B}\) be the collection of all circular regions (interiors of circles) in the plane. Then \(\mathcal{B}\) satisfies both conditions for a basis. The second condition is illustrated in Figure 13.1. In the topology generated by \(\mathcal{B}\), a subset \(U\) of the plane is open if every \(x\) in \(U\) lies in some circular region contained in \(U\).

2. Let \(\mathcal{B}^{\prime}\) be the collection of all rectangular regions (interiors of rectangles) in the plane, where the rectangles have sides parallel to the coordinate axes. Then \(\mathcal{B}^{\prime}\) satisfies both conditions for a basis. The second condition is illustrated in Figure 13.2; in this case, the condition is trivial, because the intersection of any two basis elements is itself a basis element (or empty).

Theorem 13.0

The collection \(\mathcal{T}\) generated by the basis \(\mathcal{B}\) is, in fact, a topology on \(X\).

Proof

If \(U\) is the empty set, it satisfies the defining condition of openness vacuously. Likewise, \(X\) is in \(\mathcal{T}\), since for each \(x \in X\) there is some basis element \(B\) containing \(x\) and contained in \(X\).

Now let us take an indexed family \(\left\{U_{\alpha}\right\}_{\alpha \in J}\), of elements of \(\mathcal{T}\) and show that

\[ U=\bigcup_{\alpha \in J} U_{\alpha} \]

belongs to \(\mathcal{T}\). Given \(x \in U\), there is an index \(\alpha\) such that \(x \in U_{\alpha}\). Since \(U_{\alpha}\) is open, there is a basis element \(B\) such that \(x \in B \subset U_{\alpha}\). Then \(x \in B\) and \(B \subset U\), so that \(U\) is open, by definition.

Now let us take two elements \(U_{1}\) and \(U_{2}\) of \(\mathcal{T}\) and show that \(U_{1} \cap U_{2}\) belongs to \(\mathcal{T}\). Given \(x \in U_{1} \cap U_{2}\), choose a basis element \(B_{1}\) containing \(x\) such that \(B_{1} \subset U_{1}\); choose also a basis element \(B_{2}\) containing \(x\) such that \(B_{2} \subset U_{2}\). The second condition for a basis enables us to choose a basis element \(B_{3}\) containing \(x\) such that \(B_{3} \subset B_{1} \cap B_{2}\). Then \(x \in B_{3}\) and \(B_{3} \subset U_{1} \cap U_{2}\), so \(U_{1} \cap U_{2}\) belongs to \(\mathcal{T}\), by definition.

Finally, we show by induction that any finite intersection \(U_{1} \cap \cdots \cap U_{n}\) of elements of \(\mathcal{T}\) is in \(\mathcal{T}\). This fact is trivial for \(n=1\); we suppose it true for \(n-1\) and prove it for \(n\). Now

\[ \left(U_{1} \cap \cdots \cap U_{n}\right)=\left(U_{1} \cap \cdots \cap U_{n-1}\right) \cap U_{n} . \]

By hypothesis, \(U_{1} \cap \cdots \cap U_{n-1}\) belongs to \(\mathcal{T}\); by the result just proved, the intersection of \(U_{1} \cap \cdots \cap U_{n-1}\) and \(U_{n}\) also belongs to \(\mathcal{T}\).

Thus we have checked that collection of open sets generated by a basis \(\mathcal{B}\) is, in fact, a topology.

Lemma 13.1

(Basis \(\Rightarrow\) Topology)
Let \(X\) be a set; let \(\mathcal{B}\) be a basis for a topology \(\mathcal{T}\) on \(X\). Then \(\mathcal{T}\) equals the collection of all unions of elements of \(\mathcal{B}\).

Proof

Given a collection of elements of \(\mathcal{B}\), they are also elements of \(\mathcal{T}\). Because \(\mathcal{T}\) is a topology, their union is in \(\mathcal{T}\).
Conversely, given \(U \in \mathcal{T}\), choose for each \(x \in U\) an element \(B_{x}\) of \(\mathcal{B}\) such that \(x \in B_{x} \subset U\). Then \(U=\bigcup_{x \in U} B_{x}\), so \(U\) equals a union of elements of \(\mathcal{B}\).

Lemma 13.2

(Topology \(\Rightarrow\) Basis)
Let \(X\) be a topological space. Suppose that \(C\) is a collection of open sets of \(X\) such that for each open set \(U\) of \(X\) and each \(x\) in \(U\), there is an element \(C\) of \(\mathcal{C}\) such that \(x \in C \subset U\). Then \(\mathcal{C}\) is a basis for the topology of \(X\).

Proof

We must show that \(\mathcal{C}\) is a basis. The first condition for a basis is easy: Given \(x \in X\), since \(X\) is itself an open set, there is by hypothesis an element \(C\) of \(\mathcal{C}\) such that \(x \in C \subset X\). To check the second condition, let \(x\) belong to \(C_{1} \cap C_{2}\), where \(C_{1}\) and \(C_{2}\) are elements of \(C\). Since \(C_{1}\) and \(C_{2}\) are open, so is \(C_{1} \cap C_{2}\). Therefore, there exists by hypothesis an element \(C_{3}\) in \(C\) such that \(x \in C_{3} \subset C_{1} \cap C_{2}\).

Let \(\mathcal{T}\) be the collection of open sets of \(X\); we must show that the topology \(\mathcal{T}^{\prime}\) generated by \(\mathcal{C}\) equals the topology \(\mathcal{T}\). First, note that if \(U\) belongs to \(\mathcal{T}\) and if \(x \in U\), then there is by hypothesis an element \(C\) of \(\mathcal{C}\) such that \(x \in C \subset U\). It follows that \(U\) belongs to the topology \(\mathcal{T}^{\prime}\), by definition. Conversely, if \(W\) belongs to the topology \(\mathcal{T}^{\prime}\), then \(W\) equals a union of elements of \(\mathcal{C}\), by Lemma 13.1. Since each element of \(\mathcal{C}\) belongs to \(\mathcal{T}\) and \(\mathcal{T}\) is a topology, \(W\) also belongs to \(\mathcal{T}\).

Lemma 13.3

Let \(\mathcal{B}\) and \(\mathcal{B}^{\prime}\) be bases for the topologies \(\mathcal{T}\) and \(\mathcal{T}^{\prime}\), respectively, on \(X\). Then the following are equivalent:

1. \(\mathcal{T}^{\prime}\) is finer than \(\mathcal{T}\). (\(\mathcal{T}^{\prime} \supset \mathcal{T}\))
2. For each \(x \in X\) and each basis element \(B \in \mathcal{B}\) containing \(x\), there is a basis element \(B^{\prime} \in \mathcal{B}^{\prime}\) such that \(x \in B^{\prime} \subset B\).

Proof

(2) \(\Longrightarrow\) (1).
Given an element \(U\) of \(\mathcal{T}\), we wish to show that \(U \in \mathcal{T}^{\prime}\). Let \(x \in U\). Since \(\mathcal{B}\) generates \(\mathcal{T}\), there is an element \(B \in \mathcal{B}\) such that \(x \in B \subset U\). Condition (2) tells us there exists an element \(B^{\prime} \in \mathcal{B}^{\prime}\) such that \(x \in B^{\prime} \subset B\). Then \(x \in B^{\prime} \subset U\), so \(U \in \mathcal{T}\), by definition.

(1) \(\Longrightarrow\) (2).
We are given \(x \in X\) and \(B \in \mathcal{B}\), with \(x \in B\). Now \(B\) belongs to \(\mathcal{T}\) by definition and \(\mathcal{T} \subset \mathcal{T}^{\prime}\) by condition (1); therefore, \(B \in \mathcal{T}^{\prime}\). Since \(\mathcal{T}^{\prime}\) is generated by \(\mathcal{B}^{\prime}\), there is an element \(B^{\prime} \in \mathcal{B}^{\prime}\) such that \(x \in B^{\prime} \subset B\).

Example

One can now see that the collection \(\mathcal{B}\) of all circular regions in the plane generates the same topology as the collection \(\mathcal{B}^{\prime}\) of all rectangular regions; Figure 13.4 illustrates the proof.

Definition 13.2 : Standard Topology

If \(\mathcal{B}\) is the collection of all open intervals in the real line,

\[ (a, b)=\{x \mid a<x<b\} \]

the topology generated by \(\mathcal{B}\) is called the standard topology on the real line. Whenever we consider \(\mathbb{R}\), we shall suppose it is given this topology unless we specifically state otherwise.

Definition 13.3 : Lower Limit Topology

If \(\mathcal{B}^{\prime}\) is the collection of all half-open intervals of the form

\[ [a, b)=\{x \mid a \leq x<b\} \]

where \(a<b\), the topology generated by \(\mathcal{B}^{\prime}\) is called the lower limit topology on \(\mathbb{R}\). When \(\mathbb{R}\) is given the lower limit topology, we denote it by \(\mathbb{R}_{\ell}\).

Definition 13.4 : K - Topology

Finally let \(K\) denote the set of all numbers of the form \(1 / n\), for \(n \in \mathbb{Z}_{+}\), and let \(\mathcal{B}^{\prime \prime}\) be the collection of all open intervals \((a, b)\), along with all sets of the form \((a, b)-K\).
The topology generated by \(\mathcal{B}^{\prime \prime}\) will be called the \({K}\)-topology on \(\mathbb{R}\). When \(\mathbb{R}\) is given this topology, we denote it by \(\mathbb{R}_{K}\).

Lemma 13.4

The topologies of \(\mathbb{R}_{\ell}\) and \(\mathbb{R}_{K}\) are strictly finer than the standard topology on \(\mathbb{R}\), but are not comparable with one another.

Proof

Let \(\mathcal{T}, \mathcal{T}^{\prime}\), and \(\mathcal{T}^{\prime \prime}\) be the topologies of \(\mathbb{R}, \mathbb{R}_{\ell}\), and \(\mathbb{R}_{K}\), respectively. Given a basis element \((a, b)\) for \(\mathcal{T}\) and a point \(x\) of \((a, b)\), the basis element \([x, b)\) for \(\mathcal{T}^{\prime}\) contains \(x\) and lies in \((a, b)\). On the other hand, given the basis element \([x, d)\) for \(\mathcal{T}^{\prime}\), there is no open interval \((a, b)\) that contains \(x\) and lies in \([x, d)\). Thus \(\mathcal{T}^{\prime}\) is strictly finer than \(\mathcal{T}\).

A similar argument applies to \(\mathbb{R}_{K}\). Given a basis element \((a, b)\) for \(\mathcal{T}\) and a point \(x\) of \((a, b)\), this same interval is a basis element for \(\mathcal{T}^{\prime \prime}\) that contains \(x\). On the other hand, given the basis element \(B=(-1,1)-K\) for \(\mathcal{T}^{\prime \prime}\) and the point 0 of \(B\), there is no open interval that contains 0 and lies in \(B\).

Same examples can be applied to show that \(\mathbb{R}_{\ell}\) and \(\mathbb{R}_{K}\) are not comparable with one another.

Definition 13.5 : Subbasis / Topology Generated by Subbasis

A subbasis \(S\) for a topology on \(X\) is a collection of subsets of \(X\) whose union equals \(X\). The topology generated by the subbasis \(S\) is defined to be the collection \(\mathcal{T}\) of all unions of finite intersections of elements of \(S\).

Theorem 13.5

\(\mathcal{T}\), the topology generated by the subbasis \(S\) is a topology.

Proof

For this purpose it will suffice to show that the collection \(\mathcal{B}\) of all finite intersections of elements of \(\mathcal{S}\) is a basis, for then the collection \(\mathcal{T}\) of all unions of elements of \(\mathscr{B}\) is a topology, by Lemma 13.1.

Given \(x \in X\), it belongs to an element of \(\mathcal{S}\) and hence to an element of \(\mathscr{B}\); this is the first condition for a basis.

To check the second condition, let

\[ B_{1}=S_{1} \cap \cdots \cap S_{m} \quad \text { and } \quad B_{2}=S_{1}^{\prime} \cap \cdots \cap S_{n}^{\prime} \]

be two elements of \(\mathscr{B}\). Their intersection

\[ B_{1} \cap B_{2}=\left(S_{1} \cap \cdots \cap S_{m}\right) \cap\left(S_{1}^{\prime} \cap \cdots \cap S_{n}^{\prime}\right) \]

is also a finite intersection of elements of \(\mathcal{S}\), so it belongs to \(\mathcal{B}\).