# The Euclidean topology of the real line (2)

This is another discussion of the Euclidean topology of the real line. The previous post (The Euclidean topology of the real line (1)) introduced the notion of topological spaces via a discussion of the Euclidean (usual) topology of the real number line. In this post we continue to discuss the usual topology of the real line, mostly basic facts about open sets and closed sets that we will need in subsequent posts. Also, much of what we discuss in this post about the usual topology will hold for topological spaces in general. This post is part of a series of posts on the topology on the real line. We believe that the Euclidean topology (especially the real line topology and the real plane topology) is a gateway to general topology.

This is a post on introductory topology. Concepts discussed here are basic notions found in beginning topology courses and elementary analysis courses. For the readers who are taking math courses that transition their math career from calculation to proving of theorems, we encourage such readers to think about the concepts and theorems discussed here before reading the proofs.

Definition
Let $\mathbb{R}$ be the real number line.

1. An open interval $(a,b)$ is the set of all real numbers $x$ such that $a.
2. A subset $U$ of $\mathbb{R}$ is an open set if for each $x \in U$, there is some open interval $(a,b)$ containing $x$ such that $(a,b) \subset U$.
3. The complement of $A \subset \mathbb{R}$ (denoted by $\mathbb{R}-A$) is the set of all points $x \in \mathbb{R}$ such that $x \notin A$.
4. A set $C \subset \mathbb{R}$ is a closed set if its complement $\mathbb{R}-C$ is an open set.
5. A point $p$ is said to be a limit point of the set $A \subset \mathbb{R}$ if every open set containing $p$ contains a point of $A$ different from $p$.
6. A set $A \subset \mathbb{R}$ is a perfect set if $A$ is closed and every point of $A$ is a limit point of $A$.
7. For any set $A \subset \mathbb{R}$, let $A^d$ denote the set of all limit points of $A$. The set $A^d$ is said to be the derived set of $A$.
8. The closure of  $A \subset \mathbb{R}$ is the set $\overline{A}=A \cup A^d$.
9. A set $A \subset \mathbb{R}$ is dense in $\mathbb{R}$ if $\overline{A}=\mathbb{R}$.

Since an open set of made up of open intervals, we have the following observations about open sets (stated in the following theorem).

Theorem 1

1. Both $\mathbb{R}$ and the empty set $\phi$ are open sets.
2. Let $\lbrace{U_\alpha: \alpha \in \mathcal{S}}\rbrace$ be a collection of open sets where $\mathcal{S}$ is some index set. Then the union of the open sets $U_\alpha$, denoted by $\bigcup \limits_{\alpha \in \mathcal{S}} U_\alpha$, is also an open set.
3. If $U_1,U_2, \cdots, U_n$ are open sets where $n$ is a positive integer, then $\bigcap \limits_{i=1}^n U_i$ is an open set.

Proof
The three conditions in this theorem are the axiomatic conditions for a collection of open sets to be a topology. Once we show that the open sets as defined in this post satisfy these three conditions, we demonstrate that the set of all open subsets of $\mathbb{R}$ is a topology, which is the Euclidean topology.

It is clear that $\mathbb{R}$ is open since for each point $x \in \mathbb{R}$, we have $(x-1,x+1) \subset \mathbb{R}$. Since the empty set $\phi$ has no point, it satisfies the definition of open set by default.

Let $\lbrace{U_\alpha: \alpha \in \mathcal{S}}\rbrace$ be a collection of open sets where $\mathcal{S}$ is some index set. Let $p \in \bigcup \limits_{\alpha \in \mathcal{S}} U_\alpha$. Then $p \in U_\beta$ for some $\beta \in \mathcal{S}$. By definition, there is some open interval $(a,b)$ such that $p \in (a,b) \subset U_\beta$. Note that $U_\beta \subset \bigcup \limits_{\alpha \in \mathcal{S}} U_\alpha$.

To show condition $3$, it suffices to show that the intersection of two open sets is open. Let $p \in U_1 \cap U_2$ where $U_1$ and $U_2$ are open. Then there are open intervals $(a_1,b_1)$ and $(a_2,b_2)$ such that $p \in (a_1,b_1) \subset U_1$ and $p \in (a_2,b_2) \subset U_2$. We can find a smaller open interval $(a_3,b_3)$ such that $p \in (a_3,b_3)$ and $(a_3,b_3) \subset (a_1,b_1) \cap (a_2,b_2)$. Thus we have $p \in (a_3,b_3) \subset U_1 \cap U_2$. This shows that $U_1 \cap U_2$ is an open set.

Note that we can set $(a_3,b_3)=(a_1,b_1) \cap (a_2,b_2)$ in the proof of condition $3$. In many other topological spaces, the intersection of two “open intervals” is not necessarily an “open interval”. However, we can always find an open interval contained within the intersection of two open intervals. $\blacksquare$

Corollary 2
Since a closed set is the complement of an open set, closed sets satisfy the following conditions:

1. Both $\mathbb{R}$ and the empty set $\phi$ are closed sets.
2. Let $\lbrace{C_\alpha: \alpha \in \mathcal{S}}\rbrace$ be a collection of closed sets where $\mathcal{S}$ is some index set. Then the intersection of the closed sets $C_\alpha$, denoted by $\bigcap \limits_{\alpha \in \mathcal{S}} C_\alpha$, is also a closed set.
3. If $C_1,C_2, \cdots, C_n$ are open sets where $n$ is a positive integer, then $\bigcup \limits_{i=1}^n C_i$ is a closed set.

Proposition 3

1. If the point $p$ is a limit point of the set $A \subset \mathbb{R}$, then every open set containing $p$ containing infinitely many points of $A$.
2. Finite sets of real numbers have no limit points.

Proof
Let $p \in \mathbb{R}$ be a limit point of $A \subset \mathbb{R}$. It suffices to show that every open interval containing $p$ contains infinitely many points of $A$. Let $(a,b)$ be an open interval such that $a.

Suppose that $(a,b)$ only contains finitely many points of $A$ and we aim to derive a contradiction. Suppose these finite number of points are $x_1,x_2, \cdots, x_n$. Then we take the minimum of the following differences:

$\displaystyle \lvert p-x_1 \lvert,\lvert p-x_2 \lvert, \cdots, \lvert p-x_n \lvert$

All of the above differences are positive as the points $x_i$ are different from $p$. Let $d$ be the smallest of the above $n$ differences. Consider the open interval $(p-0.1d,p+0.1d)$. Note that points in this open intervals are less than $0.1d$ away from the point $p$. The points $x_i$ are at least $d$ away from $p$. Thus we have an open interval containing $p$ that has no points of $A$ other than $p$, contradicting the fact that $p$ is a limit point of $A$. So, the assumption above that there is an open interval containing $p$ that contains only finitely many points of $A$ is faulty and the opposite must be true.

A corollary to the above is that only infinite set can have a limit point. Thus every finite set of real numbers has no limit point. $\blacksquare$

Theorem 4
Let $C \subset \mathbb{R}$. Then the following conditions are equivalent.

1. The set $C$ is a closed set.
2. If $p$ is a limit point of $C$, then $p \in C$.
3. $C=\overline{C}$.

Proof
$1 \Rightarrow 2$ Suppose $C$ is closed. Suppose we have: $p$ is a limit point of $C$ and $p \notin C$. Since $\mathbb{R}-C$ is open and $p \in \mathbb{R}-C$, $\mathbb{R}-C$ contains a point of $C$, which is impossible. Thus if $p$ is a limit point of $C$, then $p \in C$.

$2 \Leftrightarrow 3$ Condition $3$ is really just a more compact way of expressing condition $2$. Note that $\overline{C}=C \cup C^d$ where $C^d$ is the set of all limit points of $C$. Also note that $C \subset \overline{C}$. Another way of saying condition $2$ is that $\overline{C} \subset C$.

$3 \Rightarrow 1$ We prove the contrapositive: if $C$ is not closed, then $C$ has a limit point that is not in $C$. To see this, note that if $C$ is not closed, then $\mathbb{R}-C$ is not open. Then there is a point $p \in \mathbb{R}-C$ such that for each open interval $(a,b)$ containing $p$, $(a,b)$ is not a subset of $\mathbb{R}-C$ (this means $(a,b)$ contains a point of $C$). In other words, $p$ is a limit point of $C$ and $p \notin C$. $\blacksquare$

The following proposition easily follows from definition. For emphasis, we would like to call that out in a proposition.

Proposition 5
A set $D \subset \mathbb{R}$ is dense in $\mathbb{R}$ if and only if every nonempty open set of $\mathbb{R}$ contains points of $D$.

Examples
Consider the following sets:

$\displaystyle \mathbb{N}^+=\left\{n: \text{n is a positive integer}\right\}$

$\displaystyle A=\left\{\frac{1}{n}: n \in \mathbb{N}^+\right\}$

$\displaystyle B=\left\{\frac{\sqrt{2}}{10^n}: n \in \mathbb{N}^+\right\}$

$\displaystyle C=(0,1) \cup \left\{2\right\}$

$\displaystyle \mathbb{Q}=\left\{x \in \mathbb{R}: \text{x is a rational number}\right\}$

$I=[0,1]$

The set $\mathbb{N}^+$ is closed. The following is the complement of $\displaystyle \mathbb{N}^+$, which is a union of open sets.

$\displaystyle \mathbb{R}-\mathbb{N}^+=(-\infty,1) \cup (1,2) \cup (2,3) \cup \cdots$

The set $\mathbb{N}^+$ has no limit point. Note that for each $x \in \mathbb{N}^+$, $(x-0.1,x+0.1)$ is an open interval containing no point of $\mathbb{N}^+$ other than $x$. A closed set that has no limit point is said to be a discrete set.

The point $0$ is a limit point of $A$. Note that every open interval $(-c,+c)$ where $c>0$ contains $\frac{1}{n}$ for some sufficiently large positive integer $n$. None of the point in $A$ is a limit point of $A$. We have $\overline{A}=A \cup \left\{0\right\}$.

The comments made about the set $A$ can be made about the set $B$. The point $0$ is a limit point of $B$. We have $\overline{B}=B \cup \left\{0\right\}$.

The set $C$ is the union of an open interval and a single point. It is not an open set in $\mathbb{R}$ since no open interval containing the point $2$ is a subset of $C$.

The set $\mathbb{Q}$ is dense in $\mathbb{R}$. It is a property of the real line that for each pair of real numbers $a,b$ with $a there is some rational number $r$ with $a. Since $\mathbb{Q}$ is a countable set, the real line has a countable dense set. This is an important property among topological spaces. Though $\mathbb{Q}$ is not closed, every point of $\mathbb{Q}$ is a limit point of $\mathbb{Q}$.

The unit interval $I=[0,1]$ is closed as its complement is the open set $(-\infty,0) \cup (1,+\infty)$. Every point of $I=[0,1]$ is a limit point of $I$. Thus $I$ is a perfect set. A more interesting point about the unit interval is that it is a closed and bounded set and thus is a compact space. We will discuss this in a subsequent post.

Subspace Topology
Very often we need to work with a subset of $\mathbb{R}$ as a topological space. The open sets for a subset are inherited from the overall space $\mathbb{R}$. Let $Y \subset \mathbb{R}$. In working with $Y$ as a topological space, we only consider points of $Y$. Thus the open sets for $Y$ are of the form $U \cap Y$ where $U$ is an open set in the real line. We say that $U \cap Y$ is an open set relative to $Y$. As the following examples show, a set can be open in the relative topology and not open in the overall topology.

The set $\displaystyle C=(0,1) \cup \left\{2\right\}$ in the above example is clearly not an open set in $\mathbb{R}$. However, if we consider the space $C$ as a space then $C$ is an open set. Within the confine of the space $C$, the singleton set $\left\{2\right\}$ is an open set (i.e. open relative to $C$).

Let $\mathbb{P}$ be the set of all irrational numbers. Consider $\mathbb{P}$ as a topological space with open sets inherited from $\mathbb{R}$. The set $B=\left\{\frac{\sqrt{2}}{10^n}: n \in \mathbb{N}^+\right\}$ has no limit point in $\mathbb{P}$ since the point $0$ is no longer in the space. It is clear that no point of $B$ is a limit point of $B$. Thus $B$ is closed set in $\mathbb{P}$. Note that $B$ is not a }closed set in $\mathbb{R}$.

The following are two excellent texts at the introductory level.

Reference

1. Steen, L. A. and Seebach, J. A., Counterexamples in Topology, 1995, Dover Publications, Inc, New York.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.