In this section, we would like to review some interesting results about the Euclidean space $\mathbb{R}^{n}$. By a Euclidean space $\mathbb{R}^{n}$, we may mean the real inner product space $\mathbb{R}^{n}$ or the vector space $\mathbb{R}^{n}$ depending on the context. Let $U\subset\mathbb{R}^{n}$ be a subset. The Euclidean norm is given by

$$ \|x\|=\sqrt{x_{1}^{2}+...+x_{n}^{2}} $$

for any $x=(x_{1},...,x_{n})\in\mathbb{R}^{n}$ and the Euclidean distance is defined by

$$d(x,y)=\sqrt{(x_{1}-y_{1})^{2}+...+(x_{n}-y_{n})^{2}}$$

for any $x=(x_{1},...,x_{n}),y=(y_{1},...,y_{n})\in\mathbb{R}^{n}$.

Definition 1.1.

1. We say that $U$ is open if for every $x\in U$ there exists $\mathbb{B}(x,r)\subset U$, where $\mathbb{B}(x,r):=\{y\in\mathbb{R}^{n}\mid\|y-x\|<r\}$ for some $r>0$. And we say that $U$ is closed if $U^{c}$ is open.
2. An (open) neighborhood of a point $x\in\mathbb{R}^{n}$ is an open set containing $x$. A point $p\in\mathbb{R}^{n}$ is a limit point of $U$, if for every neighborhood $V$ of $p$, we have $(V-p)\cap U\neq\varnothing$. The set of all limit points of $U$ is called the derived set of $U$, which is denoted by $U'$.
3. The closure of $U$ denoted by $\overline{U}$ is the union of $U$ and all of its limits points, i.e. $\overline{U}=U\cup U'$.

Remark 1.2. By convention, the empty set is an open subset. And we would only consider open neighborhood since it will make no difference.

Proposition 1.3. Let $U\subset\mathbb{R}^{n}$ be a subset. Then the following are equivalent:

1. $U$ is a closed subset;
2. $\overline{U}=U$.

Proof. First we assume that $U$ is a closed subset, i.e. $U^{c}$ is open. Then for every $x\in U^{c}$, there exists $\mathbb{B}(x,r)\subset U^{c}$. So there is no limit point of $U$ in $U^{c}$. Consequently, we have $\overline{U}=U$ as desired.

Conversely, assume that $\overline{U}=U$, then we consider the complement $U^{c}$ of $U$. Clearly, for any $x\in U^{c}$, there exists $\mathbb{B}(x,r)\subset U^{c}$. So $U^{c}$ is open, which means $U$ is closed.

By Proposition 1.3, we can easily deduce the following proposition.

Proposition 1.4. Let $U$ be a subset of $\mathbb{R}^{n}$, the closure $\overline{U}$ of $U$ is the smallest closed set containing $U$, or equivalently, the intersection of all closed sets containing $U$.

By Definition 1.1, we can easily deduce the following proposition, which indicates the topological properties of $\mathbb{R}^{n}$.

Proposition 1.5. The following properties of $\mathbb{R}^{n}$ are satisfied:

1. $\mathbb{R}^{n},\varnothing$ are open sets;
2. Any finite intersection of open sets is open;
3. The union of an arbitrary family of open sets is open.

So we have the dual proposition about closed sets:

Proposition 1.6. The following properties of $\mathbb{R}^{n}$ are satisfied:

1. $\mathbb{R}^{n},\varnothing$ are closed sets;
2. Any finite union of closed sets is closed;
3. Arbitrary intersections of closed sets are closed.

Proposition 1.5 leads to the notion of a topology:

Definition 1.7. Let $X$ be a set and $\tau$ be a set of subsets of $X$. Then $\tau$ is called a topology on $X$, whose elements are called open sets, if it is subject to the following axioms:

1. $X,\varnothing$ are open sets;
2. Any finite intersection of open sets is open;
3. Arbitrary unions of open sets are open.

A topological space is a pair $(X,\tau)$ consisting of a set $X$ and a topology $\tau$ on $X$. By abuse of notation, we will often simply denote by $X$ when no confusion will arise.

So $\mathbb{R}^{n}$ is a topological space with the standard topology $\mathscr{O}$ defined in Proposition 1.5. And we will always hold the assumption that $\mathbb{R}^{n}$ has the standard topology $\mathscr{O}$.

Next, we discuss more about the topological properties of $\mathbb{R}^{n}$. It is observed that $\mathbb{R}^{n}$ cannot be expressible as a union of two non-empty disjoint open sets, i.e. the only subsets of $\mathbb{R}^{n}$ that are both open and closed are $\mathbb{R}^{n}$ and $\varnothing$. In fact, if we assume that $\mathbb{R}^{n}=U_{1}\cup U_{2}$ for two non-empty disjoint open sets $U_{1},U_{2}$, we readily see that $U_{1}\cup U_{2}$ cannot be equal to $\mathbb{R}^{n}$, which is a contradiction. However, disconnected regions in $\mathbb{R}^{n}$ can be expressible as a union of two non-empty disjoint open sets. We generalize this property to arbitrary topological spaces.

Definition 1.8. A topological space $X$ is connected if $X\neq U_{1}\cup U_{2}$ where $U_{1},U_{2}$ are non-empty disjoint open sets.

So $\mathbb{R}^{n}$ is a connected topological space! In fact, $\mathbb{R}^{n}$ satisfies a stronger condition.

Definition 1.9. A topological space $X$ is path-connected if any pair of points in $X$ can be joined by a curve.

Remark 1.10. To help understand, we use a more intuitive word "curve" instead of a topological terminology "path".

Clearly, $\mathbb{R}^{n}$ is a path-connected space! Next, take a subset $U$ of $\mathbb{R}^{n}$. It is natural to ask whether properties of $\mathbb{R}^{n}$ in Proposition 1.5 can be passed down to $U$. Next, we construct the subspace topology so that every subset can be equipped with a natural topology.

Definition 1.11. Let $(X,\mathscr{O})$ be a topological space and let $S\subset X$ be a subset. We define the induced or subspace topology on $S$ as follows:

$$\mathscr{O}_{S}:=\{S\cap U\mid U\in\mathscr{O}\}.$$

The topological space $(S,\mathscr{O}_{S})$ is called a subspace of $(X,\mathscr{O})$.

Remark 1.12. In the sequel, without explicitly mentioned, every subset of a topological space naturally carries the subspace topology. Note that for any closed $U\subset X$, $U\cap S$ is a closed set in the induced topology.

Then recall that in the real line case $\mathbb{R}^{1}$, every open covering consisting of open intervals of a closed interval has a finite subcovering. In fact, the $\mathbb{R}^{1}$ case can be generalized to $\mathbb{R}^{n}$ by the Heine-Borel theorem in the following.

We first introduce the concept of compactness in $\mathbb{R}^{n}$.

Definition 1.13. A subset $U\subset\mathbb{R}^{n}$ is bounded if there exists $\mathbb{B}(0,r)$ such that $\mathbb{B}(0,r)\supset U$. And we say that $U$ is compact if it is closed and bounded. Let $(U_{i})_{i\in I}$ be a family of subsets $U_{i}\subset\mathbb{R}^{n}$. Then we say that $(U_{i})_{i\in I}$ is a covering of $U$ if $U\subset\bigcup_{i\in I}U_{i}$. The covering is open if each $U_{i}$ is open. A subcovering of $(U_{i})_{i\in I}$ is a subfamily $(U_{j})_{j\in J}$ of $(U_{i})_{i\in I}$ with $J\subset I$ such that $U\subset\bigcup_{j\in J}U_{j}$.

Theorem 1.14 (The Heine-Borel theorem). A subset $U\subset\mathbb{R}^{n}$ is compact if and only if every open covering of $U$ has a finite subcovering.

This leads to the following concept of compactness of topological spaces.

Definition 1.15. Let $X$ be a topological space.

1. Let $(U_{i})_{i\in I}$ be a family of subsets $U_{i}\subset X$. Then we say that $(U_{i})_{i\in I}$ is a covering of $U$ if $U\subset\bigcup_{i\in I}U_{i}$. The covering is open if each $U_{i}$ is open in $X$. A subcovering of $(U_{i})_{i\in I}$ is a subfamily $(U_{j})_{j\in J}$ of $(U_{i})_{i\in I}$ with $J\subset I$ such that $U\subset\bigcup_{j\in J}U_{j}$. The subcovering is finite if the index set $J$ is finite.
2. The topological space $X$ is compact if every open covering of $X$ has a finite subcovering. If $S\subset X$ is a subset, then $S$ is compact if every open covering of $S$ has a finite subcovering, or equivalently, $S$ is a compact as a subspace of $X$.
3. And we say that $X$ is locally compact if every point in $X$ has a compact neighborhood.

Remark 1.16. In the sequel, by compactness, we will always mean compactness of topological spaces. Note that $S$ is a compact as a subspace of $X$ if every open covering of $S$ whose members are open in the induced topology, has a finite subcovering.

It is easy to see that $\mathbb{R}^{n}$ is locally compact.

For any pair of points $x\neq y$ in $\mathbb{R}^{n}$, we can always find $\mathbb{B}(x,r_{1})\cap\mathbb{B}(y,r_{2})=\varnothing$ for some $r_{1},r_{2}>0$, which implies separated property.

Definition 1.17. A topological space $X$ is Hausdorff if for every pair of points $x\neq y$ in $X$, there exist open sets $x\in U,y\in V$ such that $U\cap V=\varnothing$.

Remark 1.18. For any pair of points in a Hausdorff space, we can always find two disjoint neighborhoods. Clearly, $\mathbb{R}^{n}$ is a Hausdorff space.

It is observed that there is some kind of metric structure on $\mathbb{R}^{n}$, i.e. the Euclidean norm and the Euclidean distance on $\mathbb{R}^{n}$. We will generalize this property in the following section.