# Functional Analysis 1: Metric Spaces

This is the first of series of lecture notes I intend to write for a graduate Functional Analysis course I am teaching in the Fall.

What is functional analysis? Functional analysis is an abstract branch of mathematics, especially of analysis, concerned with the study of vector spaces of functions. These vector spaces of functions arise naturally when we study linear differential equations as solutions of a linear differential equation form a vector space. Functional analytic methods and results are important in various fields of mathematics (for example, differential geometry, ergodic theory, integral geometry, noncommutative geometry, partial differential equations, probability, representation theory etc.) and its applications, in particular, in economics, finance, quantum mechanics, quantum field theory, and statistical physics. Topics in this introductory functional analysis course include metric spaces, Banach spaces, Hilbert spaces, bounded linear operators, the spectral theorem, and unbounded linear operators.

While functional analysis is a branch of analysis, due to its nature linear algebra is heavily used. So, it would be a good idea to brush up on linear algebra among other things you need to study functional analysis.

In functional analysis, we study analysis on an abstract space $X$ rather than the familiar $\mathbb{R}$ or $\mathbb{C}$. In order to consider fundamental notions in analysis such as limits and convergence, we need to have distance defined on $X$ so that we can speak of nearness or closeness. A distance on $X$ can be defined as a function, called a distance function or a metric, $d: X\times X\longrightarrow\mathbb{R}^+\cup\{0\}$ satisfying the following properties:

(M1) $d(x,y)=0$ if and only if $x=y$.

(M2) $d(x,y)=d(y,x)$ (Symmetry)

(M3) $d(x,y)\leq d(x,z)+d(z,y)$ (Triangle Inequality)

Here $\mathbb{R}^+$ denotes the set of all positive real numbers. You can easily see how mathematicians came up with this definition of a metric. (M1)-(M3) are the properties that the familiar distance on $\mathbb{R}$, $d(x,y)=|x-y|$ satisfies. The space $X$ with a metric $d$ is called a metric space and we usually write it as $(X,d)$.

Example. Let $x=(\xi_1,\cdots,\xi_n), y=(\eta_1,\cdots,\eta_n)\in\mathbb{R}^n$. Define
$$d(x,y)=\sqrt{(\xi_1-\eta_1)^2+\cdots+(\xi_n-\eta_n)^2}.$$
Then $d$ is a metric on $\mathbb{R}^n$ called the Euclidean metric.

This time, let $x=(\xi_1,\cdots,\xi_n), y=(\eta_1,\cdots,\eta_n)\in\mathbb{C}^n$ and define
$$d(x,y)=\sqrt{|\xi_1-\eta_1|^2+\cdots+|\xi_n-\eta_n|^2}.$$
Then $d$ is a metric on $\mathbb{C}^n$ called the Hermitian metric. Here $|\xi_i-\eta_i|^2=(\xi_i-\eta_i)\overline{(\xi_i-\eta_i)}$.

Of course these are pretty familiar examples. If there can be only these familiar examples, there would be no point of considering abstract space. In fact, the abstraction allows to discover other examples of metrics that are not so intuitive.

Example. Let $X$ be the set of all bounded sequences of complex numbers
$$X=\{(\xi_j): \xi_j\in\mathbb{C},\ j=1,\cdots\}.$$
For $x=(\xi_j), y=(\eta_j)\in X$, define
$$d(x,y)=\sup_{j\in\mathbb{N}}|\xi_j-\eta_j|.$$
Then $d$ is a metric on $X$. The metric space $(X,d)$ is denoted by $\ell^\infty$.

Example. Let $X$ be the set of continuous real-valued functions define on the closed interval $[a,b]$. Let $x, y:[a,b]\longrightarrow\mathbb{R}$ be continuous and define
$$d(x,y)=\max_{t\in [a,b]}|x(t)-y(t)|.$$
Then $d$ is a metric on $X$. The metric space $(X.d)$ is denoted by $\mathcal{C}[a,b]$.

In a metric space $(X,d)$, nearness or closeness can be described by a neighbourhood called an $\epsilon$-ball ($\epsilon>0$) centered at $x\in X$
$$B(x,\epsilon)=\{y\in X: d(x,y)<\epsilon\}.$$
These $\epsilon$-balls form a base for the topology on $X$, called the topology on $X$ induced by the metric $d$.

Next time, we will discuss two more examples of metric spaces $\ell^p$ and $L^p$. These examples are particularly important in functional analysis as they become Banach spaces. In particular, they become Hilbert spaces when $p=2$.

# Introduction to Topology 3: Limit Points, Boundary Points, and Sequential Limits

In this lecture, we study the topological nature of some familiar notions from analysis such as limit points and limits of sequences.

Throughout this lecture, we assume that the nonempty set $S$ is a topological space.

Definition. A point $x\in S$ is called a limit point of $A\subset S$ if for any open set $U$ containing $x$, $(U-\{x\})\cap A\ne\emptyset$. The set of limiting points of $A$ is called the derived set of $A$ and is denoted by $A’$.

Exercise. If $A\subset B\subset S$, then show that $A’\subset B’$.

Definition. Let $S$ be a space. $x\in S$ is called a boundary point of $A\subset S$ if for any open set $U$ containing $x$, $U\cap A\ne\emptyset$ and $U\cap(S\setminus A)\ne\emptyset$. The set of boundary points of $A$ is called the boundary of $A$ and is denoted by $B(A)$.

In Lecture 1, we studied the notion of the closure of a set. The following theorem relates the closure, limit points and boundary points of a set.

Theorem. Let $A\subset S$. Then
$$\bar A=A\cup A’=A\cup B(A).$$

Proof. First we show that $\bar A=A\cup B(A)$. Clearly, $A\cup B(A)\subset\bar A$. If $x\not\in\bar A$. Then there exists an open set $U$ containing $x$ such that $U\cap A=\emptyset$. This implies that $x\not\in A$ and $x\not\in B(A)$.

Now we show that $\bar A=A\cup A’$. Clearly, $A\cup A’\subset\bar A$. If $x\not\in A\cup A’$, then there exists an open set $U$ containing $x$, $(U-\{x\})\cap A=\emptyset$. Since $x\not\in A$, $U\cap A=\emptyset$. Thus, $x\not\in\bar A$.

Note that $A\cup A’=A\cup B(A)$ does not necessarily mean that $A’\subset B(A)$ or $B(A)\subset A’$ as shown in the following example.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$ and let $A=(0,1)\cup\{2\}$. Then $\bar A=[0,1]\cup\{2\}$, $A’=[0,1]$, and $B(A)=\{0,1,2\}$.

Definition. Let $A,B\subset S$. $A$ is dense in $B$ if $B\subset\bar A$. We say $A$ is dense in $S$ if $\bar A=S$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Let $A=(a,b)$ and $B=[a,b]$. Then $A$ is dense in $B$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Q}}=\mathbb{R}$, $\mathbb{Q}$ is dense in $\mathbb{R}$.

Definition. $A\subset S$ is said to be nowhere dense in $S$ if $\bar A$ contains no member of $\tau\setminus\{\emptyset\}$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Z}} =\mathbb{Z}$ contains no open interval, $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$.

Exercise. If $p\geq 2$ is an integer, a $p$-adic rational is a real number $r=\frac{k}{p^n}$ for some nonnegative integer $k$ and positive integer $n$. Show that the set of $p$-adic rationals in $I=[0,1]$ is dense in $I$.

Definition. $A\subset S$ is said to be perfect if $A$ is closed and $A\subset A’$.

Exercise. The Cantor ternary set $K$ is the set of all $x\in [0,1]$ having a ternary expansion $x=\frac{t_1}{3}+\frac{t_2}{3^2}+\cdots+\frac{t_n}{3^n}+\cdots$ with $t_n\ne 1$, for all $n\in\mathbb{N}$. Intuitively, $K$ can be thought of as the set obtained from $[0,1]$ following the successive removal of all open middle thirds. Show that $K$ is uncountable, perfect, and nowhere dense in $[0,1]$.

Definition. Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $S$ and $x\in S$. We say that $\{x_n\}_{n\in\mathbb{N}}$ converges to $x$ and write $x_n\rightarrow x$ if for any open set $U$ containing $x$, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. As a sequential limit, $x$ is also denoted by $\displaystyle\lim_{n\to\infty}x_n$.

Note that a sequence may have no limits, a unique limit, or several limits, depending upon the topology on $S$.

Example. Let $\tau$ be the cofinite topology on $\mathbb{R}$ i.e.
$$\tau=\{\emptyset\}\cup\{U\subset \mathbb{R}: \mathbb{R}\setminus U\ \mbox{is finite}\}.$$
Let $x_n=n$, $n\in\mathbb{N}$. Let $x\in\mathbb{R}$ and $x\in U\in\tau$. Then since $\mathbb{R}\setminus U$ is finite, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. Hence, $x_n\rightarrow x$ for all $x\in\mathbb{R}$.

Theorem. Let $A\subset S$ and $x\in S$.

1. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ such that $x_n\rightarrow x$, then $x\in\bar A$.
2. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence of distinct points in $A$ such that $x_n\rightarrow x$, then $x\in A’$.

Proof.

1. Assume the hypothesis. Let $U$ be an open set containing $x$. Then there exists a nutural number $N$ such that $x_n\in U$ for all $n\geq N$. Since $\{x_n\}_{n\in\mathbb{N}}\subset A$, $G\cap A\ne\emptyset$. Hence, $x\in\bar A$.
2. Left as an exercise.

A limit point is not necessarily a sequential limit, and a sequential limit is not necessarily a limit point as seen in the following example.

Example. Let $S=\{a,b,c\}$ and $\tau=\{\emptyset,\{a,b\},\{c\},S\}$. Let $x_1=a$, $x_2=b$, and $x_n=c$ for all $n\geq 3$. Clearly, $x_n\rightarrow c$. However, $c$ cannot be a limit point since $c\in\{c\}\in\tau$ and $(\{c\}-\{c\})\cap\{a,b,c\}=\emptyset$. $a$ and $b$ are limit points but they cannot be sequential limits since $a,b\in\{a,b\}\in\tau$ and $x_n\notin\{a,b\}$ for all $n\geq 3$.

Exercise. Let $\tau=\{\emptyset\}\cup\{(a,\infty):a\in\mathbb{R}\}\cup\{\mathbb{R}\}$. Verify that $\tau$ is a topology on $\mathbb{R}$ and establish in $(\mathbb{R},\tau)$ the following sequential convergence and divergence:

1. If $x_n=n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\in\mathbb{R}$.
2. If $x_n=-n$, for each $n\in\mathbb{N}$, then $\{x_n\}_{n\in\mathbb{N}}$ does not converge in $\mathbb{R}$.
3. If $x_n=(-1)^n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\leq -1$.

# 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.