Topology of the Real Numbers

Abigail Huettig

19 Topology of the Real Numbers

The foundation for the discussion of the topology of $\mathbb{R}$ is the Axiom of Completeness. However, before we discuss this axiom, we must be introduced to a couple more terms, the upper bound and least upper bound of a set. Abbott provides us with the following definition [1].

Definition IV.2
A set $A\subseteq \mathbb{R}$ is bounded above if there exists a number $b\in \mathbb{R}$ such that $a\leq b$ for all $a\in A$ . The number $b$ is called an upper bound for $A$ .

An upper bound of a set $A$ in the real numbers is simply a real number that is greater than or equal to every element in $A$ . An upper bound is not required to be an element of $A$ , and it is possible for a set to have an infinite number of upper bounds. However, there is one upper bound in particular that we are usually interested in. This brings us to our next definition presented by Abbott [1].

Definition IV.3
A real number $s$ is the least upper bound for a set $A\subseteq\mathbb{R}$ if it meets the following two criteria:

$s$ is an upper bound for $A$ .
If $b$ is an upper bound for $A$ , then $s\leq b$ .

So, the least upper bound of a set $A$ , denoted as sup $A$ , is simply the smallest upper bound that exists. Now that we have defined these ideas, we are ready for the Axiom of Completeness.

Axiom of Completeness

The real numbers are the foundation on which we build the branch of mathematics known as analysis, and the Axiom of Completeness is what enables us to construct the set of real numbers from the set of rationals [1]. The defining difference between these two sets is that $\mathbb{R}$ is not filled with gaps like $\mathbb{Q}$ is, and this axiom is the formal statement of this difference [1]. Abbott gives the following statement of this axiom [1].

Axiom of Completeness. Every nonempty set of real numbers that is bounded above has a least upper bound.

This in itself does not seem very interesting. In fact, what this axiom is saying seems quite obvious. What is interesting, however, is why this axiom does not hold true for the rationals.

To explain this, Abbott uses the example of the set $S$ defined as

$S=\{r\in \mathbb{Q}:r^2<2\}.$

The least upper bound of $S$ is $\sqrt{2}$ when we are not restricted to the rationals because $\sqrt{2}$ is an irrational number [1]. If we are restricted to the rationals, then the obvious choice for the least upper bound would be the smallest rational number greater than $\sqrt{2}$ . The problem is that between any two real numbers, there exists a rational number, which is formally stated in Theorem IV.4. Whenever we think we have found the smallest rational greater than $\sqrt{2}$ , we have not because there is another rational between this supposed smallest and $\sqrt{2}$ . Thus, it is impossible to find a least upper bound when restricted to the rational numbers.

Let’s now look at an example that demonstrates how the Axiom of Completeness can be used in proving statements. The following example is a homework problem I completed for Introduction to Analysis and comes from Abbott [1].

Example 39

Let $A\subseteq\mathbb{R}$ be nonempty and bounded above, and let $s\in\mathbb{R}$ have the property that for all $n\in\mathbb{N}$ , $s+\frac{1}{n}$ is an upper bound for $A$ and $s-\frac{1}{n}$ is not an upper bound for $A$ . Show $s=\text{sup}\, A$ .

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Proof.
By the Axiom of Completeness, the least upper bound of $A$ exists. For all $n\in\mathbb{N}$ , $s+\frac{1}{n}$ cannot be the least upper bound because $s+\frac{1}{n+1}$ is a smaller upper bound. We are also given that $s-\frac{1}{n}$ is not an upper bound for all $n\in\mathbb{N}$ . Thus, the least upper bound $x$ satisfies $s-\frac{1}{n}<x<s+\frac{1}{n}$ for all $n\in\mathbb{N}$ . Therefore, $x$ must equal $s$ , and $s=\text{sup}\, A$ .

In this proof, we needed to show that the supremum (least upper bound) of $A$ actually existed before we could show what value the supremum was equal to. The Axiom of Completeness was very important in that it guaranteed the existence of the supremum of $A$ .

Using the Axiom of Completeness, we have another interesting result which is given by Abbott in the following theorem [1].

Theorem IV.4 (Density of $\mathbb{Q}$ in $\mathbb{R}$ )

For every two real numbers $a$ and $b$ with $a<b$ , there exists a rational number $r$ satisfying $a<r<b$ .

This theorem then enables us to prove the following corollary given by Abbott [1]. The proof is a homework problem I completed for Introduction to Analysis.

Corollary IV.4.1 (Density of $\mathbb{I}$ in $\mathbb{R}$ )

Given any two real numbers $a<b$ , there exists an irrational number $t$ satisfying $a<t<b$ .

Proof.
Let $a, b\in\mathbb{R}$ such that $a<b$ . Because $-\sqrt{2}\in \mathbb{R}$ and $\mathbb{R}$ is closed under addition, $a-\sqrt{2}, b-\sqrt{2}\in\mathbb{R}$ . Since $a<b$ and $\mathbb{R}$ is an ordered field,

$a-\sqrt{2}<b-\sqrt{2}.$

By Theorem IV.4, there exists $r\in\mathbb{Q}$ such that

$a-\sqrt{2}<r<b-\sqrt{2} \rightarrow a<r+\sqrt{2}<b.$

By Example 24 in chapter 11, $r+\sqrt{2}\in\mathbb{I}$ since $r\in\mathbb{Q}$ and $\sqrt{2}\in\mathbb{I}$ . Therefore, given any two real numbers $a$ and $b$ where $a<b$ , there exists an irrational number $t$ satisfying $a<t<b$ .

The density of $\mathbb{Q}$ and $\mathbb{I}$ in $\mathbb{R}$ is amazing. No matter how small the distance between two real numbers, we can always find both a rational and irrational between them. This is just incredible. But we still have much more to discuss relating to the topology of $\mathbb{R}$ so we should move on to our next interesting concepts, those of open and closed sets.

Open and Closed Sets

When discussing open and closed sets, we must be careful not to think of them in terms of the traditional meanings of open and closed. Normally, we think that when something is not open that it must be closed and vice versa. But it is possible for a set to be open, closed, neither, or both. It may also be tempting to think that a closed set is bounded and an open set is not, but that would also be incorrect thinking.

Before we get into the definitions of open and closed, we must understand a couple more ideas, epsilon neighborhoods and limit points. Abbott provides us with the following formal definitions [1].

Definition IV.5
Given a real number $a\in \mathbb{R}$ and a positive number $\epsilon>0$ , the set

$V_{\epsilon}(a)=\{x\in\mathbb{R}:|x-a|<\epsilon \}$

is called the epsilon neighborhood ( $\epsilon$ -neighborhood) of $a$ .

In other words, the $\epsilon$ -neighborhood of a real number $a$ is the set of all real numbers whose distance from $a$ is less than $\epsilon$ .

Definition IV.6
A point $x$ is a limit point of a set $A$ if every $\epsilon$ -neighborhood $V_{\epsilon}(x)$ of $x$ intersects the set $A$ at some point other than $x$ .

A limit point of a set is not necessarily an element of the set. But if $x$ is a limit point of $A$ , then at least one of the following must be true.

For all $y>x$ , either $y\in A$ or $z\in A$ such that $x<z<y$ .
For all $y<x$ , either $y\in A$ or $z\in A$ such that $x>z>y$ .

Now we are ready to introduce the definitions of open and closed sets as given by Abbott [1].

Definition IV.7
A set $O\subseteq \mathbb{R}$ is open if for all points $a\in O$ there exists an $\epsilon$ -neighborhood $V_{\epsilon}(a)\subseteq O$ .

In other words, a set $O$ is open if and only if for every element $a$ in $O$ we can find $x$ and $y$ in $O$ such that $x<a>a$ , and $|a-x|=|a-y|$ .

Definition IV.8
A set $F\subseteq \mathbb{R}$ is closed if it contains its limit points.

As mentioned previously, a limit point of a set is not necessarily an element of the set. However, if every limit point is an element, then the set is closed.

Let’s now look at a couple of examples that I wrote and completed for this paper to better understand these concepts.

Example 40

Determine whether $\mathbb{N}$ is open, closed, neither, or both.

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Solution
Consider the natural number 5. No matter what we choose $\epsilon$ to be, the $\epsilon$ -neighborhood of 5 will contain some elements that are not natural numbers. Thus, there does not exist an epsilon neighborhood of 5 that is a subset of the natural numbers, and $\mathbb{N}$ is not open. For any natural number $x$ , let $\epsilon=0.1$ . Since $x$ is the only natural number in the interval $(x-0.1,x+0.1)$ , there exists an epsilon neighborhood for every natural number $x$ that does not intersect the set of natural numbers at any point other than $x$ . Thus, every natural number is not a limit point of $\mathbb{N}$ . For $x\notin \mathbb{N}$ , we can choose $\epsilon\leq d$ , where $d$ is the distance from $x$ to the nearest natural number. So, for every $x\notin \mathbb{N}$ , there exists an epsilon neighborhood that does not contain any natural numbers and consequently, does not intersect the set of natural numbers. Thus, any $x\notin \mathbb{N}$ is not a limit point. Since there are not any limit points of $\mathbb{N}$ that are not in $\mathbb{N}$ , $\mathbb{N}$ is closed.

Example 41

Determine whether $\mathbb{R}$ is open, closed, neither, or both.

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Solution
It is clear that for every real number $x$ there exists an epsilon neighborhood of $x$ that is a subset of $\mathbb{R}$ . Thus, $\mathbb{R}$ is open. No matter what we choose $\epsilon$ to be, the $\epsilon$ -neighborhood of every real number $x$ contains other real numbers. So, every real number is a limit point of $\mathbb{R}$ . Additionally, there are no limit points of $\mathbb{R}$ not in $\mathbb{R}$ . Thus, $\mathbb{R}$ is closed. Therefore, $\mathbb{R}$ is both open and closed.

So, we have seen that sets can be open, closed, both, or neither. Now, we are going to look at another characteristic that only certain closed sets possess, compactness, along with a theorem that helps us prove a set is compact.

Heine-Borel Theorem

Before we can introduce the Heine-Borel Theorem, we need to define a couple more ideas related to sets, compactness and bounds. Both of the following definitions come from Abbott [1].

Definition IV.9
A set $A\subseteq \mathbb{R}$ is bounded if there exists $M>0$ such that $|a|\leq M$ for all $a\in A$ .

Definition IV.10
A set $K\subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$ .

While it is possible to prove a set is compact by showing that every sequence in the set has a subsequence that converges to a limit that is also in the set, we have another method we can use to prove a sequence is compact, the Heine-Borel Theorem, which is stated by Abbott as follows [1].

Theorem IV.11 (Heine-Borel Theorem)

A set $K\subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

The proof of this theorem has two parts. The first is to show that if $K$ is compact, then it is closed and bounded. I will not copy it here as it is long and can be found in Abbott [1]. The second part is to show that if $K$ is closed and bounded, then it is compact. I wrote this part of the proof as a homework problem for Introduction to Analysis. Within the proof, I will use the following theorem from Abbott [1].

Theorem IV.12

A point $x$ is a limit point of a set $A$ if and only if $x=\lim a_n$ for some sequence $(a_n)$ contained in $A$ satisfying $a_n\neq x$ for all $n\in \mathbb{N}$ .

Let’s now prove that if $K\subseteq\mathbb{R}$ is closed and bounded then it is compact.

Proof.
Because $K\subseteq\mathbb{R}$ is bounded, every sequence $(a_n)\in K$ is going to be bounded as well. By Theorem IV.19 in chapter 20, every $(a_n)$ contains a convergent subsequence $(a_{n_k})$ . So, every sequence in $K$ has a subsequence that converges to a limit. Now, we must show that for $\ell=\lim (a_{n_k})$ , $\ell\in K$ .

Because $\ell=\lim (a_{n_k})$ and $(a_{n_k})\in K$ , $\ell$ is a limit point of $K$ by Theorem IV.12. (We can assume that $(a_{n_k})\neq \ell$ for all $n\in\mathbb{N}$ because if the two were equal, then $\ell\in K$ because $(a_{n_k})\in K$ , which is what we are trying to show.) So, $\ell$ is a limit point of $K$ . Because $K$ is closed, it contains its limit points. Thus, $\ell\in K$ . Therefore, $K$ is compact.

Let’s now work through an example that makes use of the Heine-Borel Theorem. This is a homework problem I completed for Introduction to Analysis that comes from Abbott [1]. This problem also makes use of another theorem from Abbott that I will state here so that what is done in the example makes sense [1].

Theorem IV.13

The intersection of an arbitrary collection of closed sets is closed.

Example 42

Show that if $K$ is compact and $F$ is closed, then $K\cap F$ is compact.

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Proof.
Because $K$ is compact, $K$ is closed by Theorem IV.11. Because $F$ is also closed, $K\cap F$ is closed by Theorem IV.13. Because $K\cap F$ only contains elements that are in both $K$ and $F$ , $K\cap F\subseteq K$ . Because $K$ is compact, it is bounded by Theorem IV.11. That is, there exists $M>0$ such that $|k|\leq M$ for all $k\in K$ . Since every element $z\in K\cap F$ is also in $K$ , it follows that $|z|\leq M$ for all $z\in K\cap F$ . Thus, $K\cap F$ is closed and bounded. Therefore, $K\cap F$ is compact by Theorem IV.13.

This concludes our discussion of the Topology of $\mathbb{R}$ . We covered a lot of information, from the Axiom of Completeness and the density of $\mathbb{Q}$ and $\mathbb{I}$ in $\mathbb{R}$ to open and closed sets to compactness and the Heine-Borel Theorem. We even brought in some ideas connected to sequences, which we will discuss in more detail in the following section.

19 Topology of the Real Numbers

Axiom of Completeness

Open and Closed Sets

Heine-Borel Theorem

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