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}

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

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.

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  1. Pingback: Introduction to Topology 3: Limit Points, Boundary Points, and Sequential Limits | MathPhys Archive

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