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:

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

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

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

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

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