[Munkres Topology] 13. Basis for a Topology

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Definition. It $X$ 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) $\forall x\in X$, then $\exists B$ s.t. $x\in B$.

(2) $x\in B_1$, $x\in B_2$, then $\exists (x\in)B_3$ s.t. $B_3 \subset B_1 \cap B_2$.

If $\mathcal{B}$ satisfies there two conditions, then we define the topology $\mathcal{T}$ generated by $\mathcal{B}$ as follows: $U(\subset X)\in \mathcal{T}$ if $\forall x\in U$, $\exists B\in \mathcal{B}$ such that $x\in B$ and $B\subset U$, Note that each basis element is itself an element of $\mathcal{T}$.

Check.

(1) If $U=\varnothing$, $U\in \mathcal{T}$. If $U=X$, $\forall x\in U$, $\exists B$ such that $x\in B \subset U$, so $U\in \mathcal{T}$.

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(2) Now let us take an indexed family $\{ U_\alpha \}_{\alpha \in J}$, 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.

(3) 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 $B_3 \subset B_1 \cap B_2$. See Figure 13.3. 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 introduction that any finite intersection $U_1 \cap \cdots \cap U_n$ of elements of $\mathcal{T}$ is $\mathcal{T}$. This fact is trivial for $n=1$; we suppose it true for $n-1$ and prove is for $n$. Now $$(U_1 \cap \cdot \cap U_n)=(U_1 \cap \cdots \cap U_{n-1})\cap U_n$$

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 $\mathcal{T}$

Another way of describing the topology generated by a basis is given in the following lemma.

Lemma 13.1. 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 $\mathcal{B}$. $\blacksquare$

This lemma states that every open set $U$ in $X$ can be expressed as a union of basis elements. This expression for $U$ is not, however, unique. Thus the use of the term 'basis' in topology differs drastically from its use in linear algebra, where the equation expressing a given vector as a linear combination of basis vectors is unique.

We have described in two different ways how to go from a basis to the topology it generates, Sometimes we need to go in the reverse direction, from a topology to a basis generating it. Here is one way of obtaining a basis for a given topology; we shall use it frequently.

Lemma 13.2. Let $X$ be a topological space. Suppose that $\mathcal{C}$ 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 $\mathcal{U}$ 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 $\mathcal{C}$. 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 $\mathcal{C}$ 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 $\mathcal{T}'$ 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 $\mathcal{T}'$, by definition. Conversely, if $W$ belongs to the topology $\mathcal{T}'$, then $W$ equals a union of elements of $\mathcal{C}$, by the preceding lemma. Since each element of $\mathcal{C}$ belongs to $\mathcal{T}$ and $\mathcal{T}$ is a topology, $W$ also belongs to $\mathcal{T}$. $\blacksquare$

When topologies are given by basis, it is useful to have a criterion in terms of the base for determining whether one topology is finer than another. One such criterion is the following:

Lemma 13.3. Let $\mathcal{B}$ and $\mathcal{B}'$ be base for topologies $\mathcal{T}$ and $\mathcal{T}'$, respectively, on $X$. Then the following are equivalent:

(1) $\mathcal{T}'$ is finer than $\mathcal{T}$.

(2) For each $x\in X$ and each basis element $B\in \mathcal{B}$ containing $x$, there is a basis element $B' \in \mathcal{B}'$ such that $x\in B' \subset B$.

Proof. $(2)\rightarrow (1).$ Given an element $U$ of $\mathcal{T}$, we wish to show that $U\in \mathcal{T}'$. 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 $B' \in \mathcal{B}'$ such that $x\in B'\subset B$. Then $x\in B'\subset U$, so $U\in \mathcal{T}$, be definition.

$(1)\rightarrow (2).$ We are given $x\in X$ and $B\in \mathcal{B}$. Now $B$ belongs to $\mathcal{T}$ by definition and $\mathcal{T} \subset \mathcal{T}'$ by condition (1); therefore, $B\in \mathcal{T}'$. Since $\mathcal{T}'$ is generated by $\mathcal{B}'$, there is an element $B' \in \mathcal{B}'$ such that $x\in B' \subset B$. $\blacksquare$

Definition. If $\mathcal{B}$ is the collection of all open intervals in the real line, $$(a,b)=\{x | a<x<b\},$$ the topology generated by $\mathcal{B}$ is called the standard topology on the real line. Whenever we consider $\mathcal{R}$, we shall suppose it is given this topology unless we specifically state otherwise. If $\mathcal{B}'$ is the collection of all half-open intervals of the form $$[a,b)=\{ x| a \le x < b\},$$ where$a<b$, the topology generated by $\mathcal{B}'$ is called the lower limit topology on $\mathbb{R}$. When $\mathbb{R}$ is given the lower limit topology, we denote it by $\mathbb{R}_l$. Finally let $K$ denote the set of all numbers of the form $1/n$, for $n\in \mathbb{Z}_+$, and let $\mathcal{B}''$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. The topology generated by $\mathcal{B}''$ will called the $K$-topology on $\mathbb{R}$. When $\mathbb{R}$ is given this topology, we denote it by $\mathbb{R}_K$.

It is easy to see that all three of these collections are bases; in each case, the intersection of two basis elements is either another basis element or is empty. The relation between these topologies is the following:

Lemma 13.4. The topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are strictly finer than the standard topology on $\mathbb{R}$, but are not comparable with one another.

Proof. Let $\mathcal{T}$, $\mathcal{T}'$, and $\mathcal{T}''$ be the topologies of $\mathbb{R}$, $\mathbb{R}_l$, 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 $\mathcal{T}'$ contains $x$ and lies in $(a,b)$. On the other hand, given the basis element $[x,d)$ for $\mathcal{T}'$, there is no open interval $(a,b)$ that contains $x$ and lies in $[x,d)$. Thus $\mathcal{T}'$ 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 $\mathcal{T}''$ that contains $x$. On the other hand, given the basis element $B=(-1,1)-K$ for $\mathcal{T}''$ and the point $0$ of $B$, there is no open interval that contains $0$ and lies in $B$.

Topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are not comparable. $\blacksquare$

A question may occur to you at this point. Since the topology generated by a basis $\mathcal{B}$ may be described as the collection of arbitrary unions of elements of $\mathcal{B}$, what happens if you start with a given collection of sets and take finite intersections of them as well as arbitrary unions? This question leads to the notion of a subbasis for a topology.

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

We must of course check that $\mathcal{T}$ is a topology. 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 $\mathcal{B}$ 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 $\mathcal{B}$; this is the first condition for a basis. To check the second condition, let $$B_1=S_1 \cap \cdots \cap S_m\quad \mbox{and}\quad B_2=S'_1\cap \cdots \cap S'_n$$ be two elements of $\mathcal{B}$. Their intersection $$B_1 \cap B_2 = (S_1 \cap \cdots \cap S_m)\cap (S'_1 \cap \cdots \cap S'_n)$$ is also a finite intersection of elements of $\mathcal{S}$, so it belongs to $\mathcal{B}$.