Monthly Archives: January 2013

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 $\mathbb{R}$.

Introduction to Topology 1: Open and Closed Sets

In the first lecture, we study open sets and closed sets which are the building blocks of topology. Let us begin with the definition of open sets and topology.

Definition. Let $S$ be a nonempty set and $\tau\subset 2^S$ such that

O1. $\emptyset, S\in\tau$.

O2. If $U_\alpha\in\tau$ for each $\alpha\in\Lambda$, then $\bigcup_{\alpha\in\Lambda}
U_\alpha\in\tau$.

O3. If $U_i\in\tau$ for each $i=1,\cdots,n$, then $\bigcap_{i=1}^nU_i\in\tau$.

Then $\tau$ is called a topology on $S$, the elements of $\tau$ are called open sets, and the ordered pair $(S,\tau)$ is called a topological space or simply a space.

Example. Let $S=\{a,b\}$. Then there are four possible topologies on $S$. They are
\begin{align*}
\tau_1&=\{\emptyset,S\},\\
\tau_2&=\{\emptyset,\{a\},S\},\\
\tau_3&=\{\emptyset,\{b\},S\},\\
\tau_4&=\{\emptyset,\{a\},\{b\},S\}.
\end{align*}

Exercise. Let $S=\{a,b,c\}$. Find all possible topologies on $S$. There are exactly 29 of them.

Definition. Let $S\ne\emptyset$. The smallest topology on $S$ is $\{\emptyset,S\}$ and is called the indiscrite topology. A space with the indiscrete topology is called an indiscrete space. The largest topology on $S$ is the power set $2^S$ and is called the discrete topology. A space with the discrete topology is called a discrete space.

Definition. Let $(S,\tau)$ be a space and $A\subset S$. $A$ is said to be closed if its compliment is open i.e. $A^c=S\setminus A\in\tau$.

Theorem. Let $(S,\tau)$ be a space. Then

C1. $\emptyset,S$ are closed.

C2. If $A_\alpha\subset S$ is closed for each $\alpha\in\Lambda$, then $\bigcap_{\alpha\in\Lambda}A_\alpha$ is closed.

C3. If $A_i\subset S$ is closed for each $i=1,\cdots,n$, then $\bigcup_{i=1}^nA_i$ is closed.

Proof. C1 is trivial. C2 and C3 can be easily shown by using De Morgan’s laws.

Remark. One may also define a topology using closed sets instead of open sets. Let $S\ne\emptyset$. Let $\mathcal{F}\subset 2^S$ satisfying C1, C2, C3:

C1. $\emptyset,S\in\mathcal{F}$.

C2. If $A_\alpha\in\mathcal{F}$ for each $\alpha\in\Lambda$, then $\bigcap_{\alpha\in\Lambda}A_\alpha\in\mathcal{F}$.

C3. If $A_i\subset S\in\mathcal{F}$ for each $i=1,\cdots,n$, then $\bigcup_{i=1}^nA_i\in\mathcal{F}$.

Let $\tau=\{U\subset S: S\setminus U\in\mathcal{F}\}$. Then $\tau$ is a topology on $S$.

Definition. Let $(S,\tau)$ be a space and $A\subset S$. The closure of $A$ is the smallest closed set containing $A$, that is
$$\bar A=\{F:A\subset F\ \mbox{and}\ F\ \mbox{is closed}\}.$$
Clearly $A\subset S$ is closed in $S$ if and only if $A=\bar A$.

Theorem. Let $(S,\tau)$ be a space and $A\subset S$. Then $x\in\bar A$ if and only if for any open set $U$ containing $x$, $U\cap A\ne\emptyset$.

Proof. Let $x\in\bar A$ and $U$ be an open set containing $x$. Suppose that $U\cap A=\emptyset$. Then $A\subset S\setminus U$ and $S\setminus U$ is closed. Since $\bar A$ is the smallest closed set containing $A$, $x\in\bar A\subset S\setminus U$. This is a contradiction. Hence $U\cap A\ne \emptyset$.

Assume that for any open set $U$ containing $x$, $U\cap A\ne \emptyset$. If $x\not\in\bar A$, then $x\in S\setminus\bar A=\cup\{S\setminus F: A\subset F\ \mbox{and}\ F\ \mbox{is closed}\}$. So there is a closed set $F$ containing $A$ such that $x\in S\setminus F$. Since $S\setminus F$ is open, this is a contradiction. Hence, $x\in\bar A$.

Exercise. Let $A,B$ be subsets of a space. Show that if $A\subset B$ then $\bar A\subset\bar B$.

Exercise. Let $A,B$ be subsets of a space. Prove or disprove:

  1.  $\overline{A\cup B}=\bar A\cup\bar B$.
  2. $\overline{A\cap B}=\bar A\cap\bar B$.

Exercise. Let $(S,\tau)$ be a space and $A_\alpha\in 2^S$ for each $\alpha\in\Lambda$. Show that

  1. $\overline{\bigcap_{\alpha\in\Lambda}A_\alpha}\subset\bigcap_{\alpha\in\Lambda}\bar A_\alpha$.
  2. $\overline{\bigcup_{\alpha\in\Lambda}A_\alpha}\supset\bigcup_{\alpha\in\Lambda}\bar A_\alpha$.

Give examples to show that the inclusions in 1 and 2 cannot be replaced by equality.