Introduction to Topology 2: Bases and Subbases

In this lecture, we study on how to generate a topology on a set from a family of subsets of the set. The idea is pretty much similar to basis of a vector space in linear algebra.

Definition. Let $(S,\tau)$ be a space and $\mathcal{B}\subset 2^S$. $\mathcal{B}$ is called a base for $\tau$ if
$$\tau=\{\emptyset\}\cup\left\{U\subset S:U=\bigcup_{\alpha\in\Lambda}B_\alpha,B_\alpha\in\mathcal{B},\forall\alpha\in\Lambda\right\}.$$
Clearly $\tau$ is base for itself.

Theorem 1. Let $S\ne \emptyset$ and $\mathcal{B}\subset 2^S$. $\mathcal{B}$ is a base for a topology on $S$ if and only if

1. $S=\bigcup\{B:B\in\mathcal{B}\}$.

2. For any $B_1,B_2\in\mathcal{B}$ with $x\in B_1\cap B_2$, there exists $B_3\in\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2$.

Proof. Suppose that $\mathcal{B}$ is a base for a topology $\tau$ on $S$. Since $S\in\tau$, it should be a union of members of $\mathcal{B}$, so $S\subset\bigcup\{B:B\in\mathcal{B}\}$ and hence $S=\bigcup\{B:B\in\mathcal{B}\}$. Let $B_1,B_2\in\mathcal{B}$ with $x\in B_1\cap B_2$. Since $B_1,B_2\in\tau$, $B_1\cap B_2\in\tau$. So $B_1\cap B_2=\bigcup_{\alpha\in\Lambda}B_\alpha$ where $B_\alpha\in\mathcal{B}$ for $\alpha\in\Lambda$. Since $x\in B_1\cap B_2$, $x\in B_\alpha$ for some $\alpha\in\Lambda$. Set $B_3=B_\alpha$. Then we are done.

Suppose that 1 and 2 hold. Let
$$\tau=\{\emptyset\}\cup\left\{U\subset S:U=\bigcup_{\alpha\in\Lambda}B_\alpha,B_\alpha\in\mathcal{B},\forall\alpha\in\Lambda\right\}.$$ Then clearly $\emptyset,S\in\tau$. So O1 is satisified. Let $U_\alpha\in\tau$, $\alpha\in\Lambda$. For each $x\in\bigcup_{\alpha\in\Lambda}U_\alpha$, there exist $B_x\in\mathcal{B}$ such that $x\in B_x\subset\bigcup_{\alpha\in\Lambda}U_\alpha$, so $\bigcup_{\alpha\in\Lambda}U_\alpha=\bigcup_{x\in \bigcup_{\alpha\in\Lambda}U_\alpha}B_x\in\tau$. Hence O2 is satisfied. Let $U_1,U_2\in\tau$ and $x\in U_1\cap U_2$. Then there exist $B_1,B_2\in\mathcal{B}$ such that $x\in B_1\subset U_1$ and $x\in B_2\subset U_2$. Since $x\in B_1\cap B_2$, by property 2 there exists $B_3\in\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2\subset U_1\cap U_2$. Thus $G_1\cap G_2$ can be expressed as a union of members of $\mathcal{B}$. Hence $U_1\cap U_2\in\tau$ and O3 is satisfied. Therefore $\mathcal{B}$ is a base for a topology $\tau$ on $S$.

Example. Let $\mathbb{R}$ be the set of real numbers and $\mathcal{B}_1=\{(a,b): a,b\in\mathbb{R}, a<b\}$. Then $\mathcal{B}_1$ is a base for the Euclidean topology (also called the interval topology) $\xi$ on $\mathbb{R}$.

Exercise: Prove that $\mathcal{B}_1$ is a base for a topology.

Example. Let $\mathcal{B}_2=\{[a,b): a,b\in\mathbb{R}, a<b\}$. Then $\mathcal{B}_2$ is a base for a topology $\mathcal{L}$ on $\mathbb{R}$. $\mathcal{L}$ is called the lower limit topology on $\mathbb{R}$. Similarly $\mathcal{B}_3=\{(a,b]:a,b\in\mathbb{R}, a<b\}$ is a base for a topology $\mathcal{U}$ on $\mathbb{R}$ called the upper limit topology. $(\mathbb{R},\mathcal{L})$ and $(\mathbb{R},\mathcal{U})$ have the same topological properties. However, $(\mathbb{R},\xi)$ and $(\mathbb{R},\mathcal{L})$ have different topological properties.

Exercise: Prove that $\mathcal{B}_2$ is a base for a topology.

Proposition. Let $(S,\tau)$ be a space and $\mathcal{B}$ a base for $\tau$. Let $U\subset S$. Then $U\in\tau$ if and only if for any $x\in U$, there exists $B\in\mathcal{B}$ such that $x\in B\subset U$.

Proof. Left as an exercise for the readers.

Definition. Let $\mathcal{B}_1,\mathcal{B}_2\subset 2^S$. Then $\mathcal{B}_1$ and $\mathcal{B}_2$ are said to be equivalent bases if they are bases for the same topology on $S$. If $\mathcal{B}_1$ and $\mathcal{B}_2$ are equivalent bases, we write $\mathcal{B}_1\sim\mathcal{B}_2$.

Theorem. Let $(S,\tau)$ be a space. Let $\mathcal{B}_1$ be a base for $\tau$ and $\mathcal{B}_2\subset 2^S$. Suppose that the following conditions 1 and 2 are satisfied:

1. If $x\in B_1\in\mathcal{B}_1$, then there exists $B_2\in\mathcal{B}_2$ such that $x\in B_2\subset B_1$.

2. If $x\in B_2\in\mathcal{B}_2$, then there exists $B_1\in\mathcal{B}_1$ such that $x\in B_1\subset B_2$.

Then $\mathcal{B}_2$ is also a base for $\tau$ and hence $\mathcal{B}_1\sim\mathcal{B}_2$.

Proof. Let $U$ be a nonempty open set. Then $U=\bigcup_{\alpha\in\Lambda}B_\alpha$, $B_\alpha\in\mathcal{B}_1$, $\alpha\in\Lambda$. If $x\in U$ then $x\in B_\alpha$ for some $\alpha\in\Lambda$. By condition 1, there exists $B\in\mathcal{B}_2$ such that $x\in B\subset B_\alpha\subset U$. So for any $x\in U$ there exists $B_x\in\mathcal{B}_2$ such that $x\in B_x\subset U$, and hence $U=\bigcup_{x\in U}B_x$. Conversely, if $U$ is a union of members of $\mathcal{B}_2$, then by condition 2 $U$ is expressed as a union of members of $\mathcal{B}_1$, and so $U\in\tau$. Therefore, $\mathcal{B}_2$ is also a base for the topology $\tau$.

Example. Consider $\mathbb{R}^2$. The (Euclidean) distance between $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb{R}^2$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. For any $(x_0,y_0)\in\mathbb{R}^2$ and $\epsilon>0$, let
$$B((x_0,y_0);\epsilon):=\{(x,y)\in\mathbb{R}^2: \sqrt{(x-x_0)^2+(y-y_0)^2}<\epsilon\}.$$ The collection
$$\mathcal{B}_1=\{B((x_0,y_0);\epsilon): (x_0,y_0)\in\mathbb{R}^2, \epsilon>0\}$$
is a base for a topology $\xi^2$ on $\mathbb{R}^2$, called the Euclidean topology on $\mathbb{R}^2$. An equivalent base is
$$\mathcal{B}_2=\{\{(x,y)\in\mathbb{R}^2: a<x<b, c<y<d\}:a,b,c,d\in\mathbb{R}\},$$
the collection of all open rectangular regions in the plane $\mathbb{R}^2$.

Exercise: Prove that $\mathcal{B}_1$ and $\mathcal{B}_2$ are equivalent bases for a topology on $\mathbb{R}^2$.

Definition. Let $(S,\tau)$ be a space and $\mathcal{S}\subset 2^S$. Let
$$\mathcal{B}=\{F_1\cap\cdots\cap F_n: F_1,\cdots,F_n\in\mathcal{S}\}$$
i.e. the collection of finite intersections of members of $\mathcal{S}$. $\mathcal{S}$ is called a subbase for $\tau$ if $\mathcal{B}$ is a base for $\tau$.

Theorem. If $\mathcal{S}\subset 2^S$ and $\bigcup\{U:U\in\mathcal{S}\}=S$, then $\mathcal{S}$ is a subbase for a unique topology $\tau$ on $S$.

Proof. We prove this theorem using Theorem 1. Let $\mathcal{B}=\{F_1\cap\cdots\cap F_n: F_1,\cdots,F_n\in\mathcal{S}\}$. Clearly $\mathcal{S}\subset\mathcal{B}$, so $\mathcal{B}\ne\emptyset$. Let $x\in S$. Since $\bigcup\{U:U\in\mathcal{S}\}=S$, there exists $U\in\mathcal{S}$ such that $x\in U$. Since $U\in\mathcal{B}$, $x\in\bigcup\{B:B\in\mathcal{B}\}$, so $S\subset \bigcup\{B:B\in\mathcal{B}\}$ i.e. the condition 1 in Theorem 1 is satisfied. Let $B_1,B_1\in\mathcal{B}$ with $x\in B_1\cap B_2$. Then $B_1=F_1\cap\cdots\cap F_m$ and $B_2=G_1\cap\cdots\cap G_n$ for some $F_1,\cdots,F_m,G_1,\cdots,G_n\in\mathcal{S}$. Put $B_3=B_1\cap B_2$. Then $B_3\in\mathcal{B}$ and $x\in B_3$. Hence the condition 2 in Theorem 1 is also satisfied. This completes the proof.

Example. Let $\mathcal{S}=\{(-\infty,b): b\in\mathbb{R}\}\cup\{(a,\infty):a\in\mathbb{R}\}$. Then $\mathcal{S}$ is a subbase for the Euclidean topology $\xi$ on $\mathbb{R}$.

Exercise. Prove that $\mathcal{S}$ in the above example is a subbase for the Euclidean topology $\xi$ on $\mathbb{R}$.

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