# The Euclidean topology of the real line (3) – Completeness

We discuss the completeness of the real line. With respect to the real line, completeness refers to the fact that the real line has no holes or gaps. There are many ways to state what it means for the real line to be complete. In this post, we state it using the least upper bound property. Then we show that this property is equivalent to a host of other statements. This discussion will be useful for subsequent posts where we discuss compactness and other properties of the real line.

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.

Let $A \subset \mathbb{R}$. By an upper bound of the set $A$, we mean a real number number $w$ such that $x \le w$ for all $x \in A$. If a set has an upper bound, we say that it is bounded above. Similarly, a lower bound of the set $A$ is a real number $y$ such that $y \le x$ for all $x \in A$. If a set has a lower bound, then we say it is bounded below. If a set is both bounded above and bounded below, it is said to be bounded.

Suppose $A \subset \mathbb{R}$ is bounded above. By a least upper bound of the set $A$, we mean a real number $u$ that is the smallest among all the upper bounds of $A$, that is, the number $u$ is itself an upper bound of $A$ and for each upper bound $w$ of $A$, we have $u \le w$. It turns out that if a set has a least upper bound, then it is unique. Suppose $u$ and $v$ are both least upper bounds of $A$. Then we have $u \le v$ and $v \le u$. This imples $u=v$.

Similarly, the number $z$ is the greatest lower bound of the set $A \subset \mathbb{R}$ if the number $z$ is a lower bound of $A$ and for each lower bound $y$ of $A$, we have $y \le z$. As with the least upper bound, if a set has a greatest lower bound, it has only one greatest lower bound. The following is the statement of the least upper bound property:

The Least Upper Bound Property. Every subset of the real line that is bounded above has a least upper bound.

In the discussion about the real line in this post and in subsequent posts, we will take the least upper bound property as a given. In the remainder of this post, we show that this property is equivalent to several other statements about the real line. Thus we could have taken any one of those statements as a given. We also have a statement that can be called the greatest lower bound property: every subset of the real line that is bounded below has a greatest lower bound. This statement is also equivalent to the least upper bound property.

For the definition of open intervals, open sets and closed sets, see the previous post The Euclidean topology of the real line (2). We have some more definitions.

Definitions

1. Let $A \subset \mathbb{R}$. Let $p \in \mathbb{R}$. The point $p$ is a limit point of $A$ if every open set containing $p$ contains a point of $A$ different from $p$.
2. Let $\mathbb{N}$ be the set of all positive integers. A sequence of real numbers is a function $f:\mathbb{N} \rightarrow \mathbb{R}$. We usually express $f(n)$ as $f_n$ and the function $f$ is expressed as $\lbrace{f_n}\rbrace_{n \in \mathbb{N}}$ or $\lbrace{f_n}\rbrace$.
3. Given a sequence of real numbers $\lbrace{f_n}\rbrace$, consider an infinite subset $S \subset \mathbb{N}$. If we only consider terms of the sequence $\lbrace{f_n}\rbrace$ where the subscript $n \in S$, then it is called a subsequence of $\lbrace{f_n}\rbrace$. The notation is $\lbrace{f_n}\rbrace_{n \in S}$.
4. A sequence $\lbrace{f_n}\rbrace$ converges if there is a number $p$ such that for every open interval $(a,b)$ containing $p$, there is a positive integer $m$ such that $f_n \in (a,b)$ for all $n \ge m$. The point $p$ is said to be the limit of $\lbrace{f_n}\rbrace$. We use the notation $\lim \limits_{n \rightarrow \infty}f_n=p$ or $f_n \rightarrow p$.
5. A sequence $\lbrace{f_n}\rbrace$ is said to be a Cauchy sequence if for each $c>0$, there is a positive integer $m$ such that $\lvert f_i-f_j \lvert for all $i,j \ge m$.
6. A sequence $\lbrace{f_n}\rbrace$ is said to be bounded if the set of the terms of the sequence $\left\{f_n: n \in \mathbb{N}\right\}$ is a bounded set, or equivalently there is a positive number $M$ such that for each term $f_n$ we have $\lvert f_n \lvert \le M$.
7. A sequence $\lbrace{f_n}\rbrace$ is said to be nondecreasing if $f_n \le f_{n+1}$ for each $n \in \mathbb{N}$. For such a sequence, we have $f_1 \le f_2 \le f_3 \le \cdots$.
8. A sequence $\lbrace{f_n}\rbrace$ is said to be nonincreasing if $f_n \ge f_{n+1}$ for each $n \in \mathbb{N}$. For such a sequence, we have $f_1 \ge f_2 \ge f_3 \ge \cdots$.
9. A sequence $\lbrace{f_n}\rbrace$ is said to be monotonic if it is either nondecreasing or nonincreasing.

One of the properties that is equivalent to the least upper bound property is the nested interval property. It is stated as follows:

The Nested Interval Property. Suppose that for each positive interger $n$ we have closed intervals $I_n=[a_n,b_n]$ such that

$\displaystyle [a_1,b_1] \supset [a_2,b_2] \supset [a_3,b_3] \supset \cdots$

and $\lim \limits_{n \rightarrow \infty}(b_n-a_n)=0$.

Then the intersection of the intervals $I_n$ has only one point, i.e. $\bigcap \limits_{n=1}^{\infty} [a_n,b_n]=\left\{x\right\}$.

We incorporate some observations about sequences into a lemma, which is used in proving the main theorem below.

Lemma

1. If a sequence $\lbrace{x_n}\rbrace$ of real numbers converges, then the limit is unique.
2. If a sequence $\lbrace{x_n}\rbrace$ of real numbers converges, then the sequence is a Cauchy sequence.
3. If $\lbrace{x_n}\rbrace$ is a Cauchy sequence of real numbers, then it is bounded.
4. If a sequence $\lbrace{x_n}\rbrace$ of real numbers converges, then the sequence is bounded.
5. If the real number $p$ is a limit point of the set $A \subset \mathbb{R}$, then there is a sequence $\lbrace{x_n}\rbrace$ of points of $A$ such that $\lbrace{x_n}\rbrace$ converges to $p$.

Proof
$(1)$ Note that if a sequence converges to a point $p$, then every open interval containing $p$ would contain all but finitely many terms in the sequence. Point $(1)$ in the lemma stems from the fact that every pair of distinct numbers in the real line are contained in two disjoint open intervals. For $x, we have disjoint open intervals $(x-0.1c,x+0.1c)$ and $(y-0.1c,y+0.1c)$ where $c=y-x$. If both $x$ and $y$ are limits of the same sequence, then each of these two disjoint open intervals would contain all but finitely many terms in the sequence, which is impossible.

$(2)$ Let $\lbrace{x_n}\rbrace$ be a convergent sequence with $p$ as a limit. For each $c>0$, we have $m$ such that $x_n \in (p-0.5c,p+0.5c)$ for all $n \ge m$. Thus for all $i,j \ge m$, we have:

$\displaystyle \lvert x_i-x_j \lvert \le \lvert x_i-p \lvert + \lvert p-x_j \lvert < 0.5c+0.5c=c$

The above shows that the sequence is a Cauchy sequence.

$(3)$ Let $\lbrace{x_n}\rbrace$ be a Cauchy sequence. For $c=0.5$, there is some positive integer $m$ such that for all $i,j \ge m$, we have $\lvert x_i-x_j \lvert<0.5$. Let $w$ be the maximum of the following real numbers:

$0.5, \lvert x_1-x_{m} \lvert, \lvert x_2-x_{m} \lvert, \cdots, \lvert x_{m-1}-x_{m} \lvert$

Then for each term $x_n$ in the sequence we have $\lvert x_n-x_m \lvert \le w$. Furthermore, we have:

$\displaystyle \lvert x_n \lvert=\lvert x_n-x_m+x_m \lvert \le \lvert x_n-x_m \lvert+\lvert x_m \lvert \le w+ \lvert x_m \lvert$

Thus $w+ \lvert x_m \lvert$ is an upper bound of the cauchy sequence in question.

$(4)$ This follows from $(2)$ and $(3)$.

$(5)$ Let $p$ be a limit point of the set $A \subset \mathbb{R}$. For each positive integer $j$, let $x_j \in A$ such that $x_j \in (p-\frac{1}{j},p+\frac{1}{j})$. Then $x_j \rightarrow p$.

Theorem
The following statements are equivalent:

1. The least upper bound property.
2. Every bounded montonic sequence of real numbers converges.
3. The nested interval property.
4. (Bolzano-Weierstrass Theorem) Every infinite set of real numbers that is also bounded has a limit point.
5. Every bounded sequence of real numbers has a subsequence that converges.
6. Every Cauchy sequence of real numbers converges.

Remark
In some texts such as $[1]$, completeness is defined in terms of Cauchy sequences (condition $6$). In a metric space, a Cauchy sequence is one where the distances of the successive terms diminishes to zero. Any space with a metric such that every Cauchy sequence converges is said to be a complete metric space. Thus the real line with the Euclidean metric is an example of a complete metric space.

In some texts, the nested interval property is taken as the definition of completeness. In a general metric space, the nested intervals are taken to be closed and bounded sets (bounded by the metric in question) with diameters $\rightarrow 0$.

The least upper bound property is based on an order relation and is not applicable to metric spaces in general. Conditions $1,2,4,5$ do not hold in metric spaces in general but hold in Euclidean spaces.

Proof
We establish the equivalences by proving $(1) \rightarrow (2)$, $(2) \rightarrow (3)$, $(3) \rightarrow (4)$, $(4) \rightarrow (5)$, $(5) \rightarrow (6)$, $(6) \rightarrow (3)$ and $(3) \rightarrow (1)$.

$(1) \rightarrow (2)$
We show that every bounded nondecreasing sequence $\lbrace{x_n}\rbrace$ of real numbers converges. For the proof that every bounded nonincreasing sequence of real numbers converges, use a similar proof by working with the greatest lower bound. Suppose that the sequence $\lbrace{x_n}\rbrace$ is nondecreasing and bounded. Let $u$ be an upper bound of the set $A=\left\{x_1,x_2,x_3, \cdots\right\}$. By the least upper bound property, the set $A$ has a least upper bound $v$. We have the following:

$\displaystyle x_1 \le x_2 \le x_3 \le \cdots \le v \le u$

We claim that the number $v$ is the limit of the sequence $\lbrace{x_n}\rbrace$. Let $(a,b)$ be an open interval containing $v$, i.e, $a. There must be some term $x_j$ of the sequence such that $a. Otherwise, $a$ would be an upper bound, which is impossible since $v$ is the least upper bound. Since the sequence is nondecreasing, we have $a for all $k \ge j$. Thus $(2)$ is established.

$(2) \rightarrow (3)$
Suppose we have a sequence of nested intervals $[a_n,b_n]$ such that the widths of the intervals $b_n-a_n \rightarrow 0$. The sequence $\lbrace{a_n}\rbrace$ is nondecreasing and has an upper bound (e.g. $b_1$). By $(2)$, this sequence converges, say to the limit $a$. On the other hand, the sequence $\lbrace{b_n}\rbrace$ is nonincreasing and has an lower bound (e.g. $a_1$). Thus this sequence converges, say to the limit $b$.

Note that $[a,b] \subset [a_n,b_n]$ for each $n$. Since $b_n-a_n \rightarrow 0$, $a=b$.

$(3) \rightarrow (4)$
Suppose $A \subset \mathbb{R}$ is an infinite set and is bounded. We produce a limit point by the approach of “divide and conquer”. Suppose that $A \subset [a,b]$. Let $a_1=a$ and $b_1=b$. Divide this interval into two equal halves, $[a_1,m]$ and $[m,b_1]$ where $m=0.5(a_1+b_1)$. Since $A$ is infinite, one of these two halves has to contain infinitely many points of $A$. Otherwise, $A$ is a finite set. Pick a half that has infinitely many points of $A$ and let $a_2$ be the left endpoint of this interval and $b_2$ be the right endpoint.

Continue this process inductively. Suppose $[a_n,b_n]$ has been chosen. Then we divide it into $[a_n,m]$ and $[m,b_n]$ where $m=0.5(a_n+b_n)$. Pick a half that has infinitely many points of $A$ and let $a_{n+1}$ be the left endpoint and $b_{n+1}$ be the right endpoint. Then we have a nested sequence of decreasing intervals. Since we keep dividing each stage into halves, the lengths of the intervals $b_n-a_n$ approaches $0$. By the nested interval property, there is only one point $x$ in the intersection of the intervals.

The point $x$ is a limit point of $A$. To see this, let $(s,t)$ be an open interval containing $x$. Then for some sufficiently large $n$ we have $[a_n,b_n] \subset (s,t)$. Then the open interval $(s,t)$ contains infinitely many points of $A$.

$(4) \rightarrow (5)$
Suppose $\lbrace{x_n}\rbrace$ is a bounded sequence of real numbers. We aim to produce a subsequence that converges. Let $A$ be the set of all terms in the sequence. The set $A$ is expressed as the following:

$A=\left\{x_1,x_2,x_3, \cdots \right\}$

Suppose $A$ is a finite set. Then starting at some integer $m$, $x_m=x_{m+1}=x_{m+2}=\cdots$. The sequence $\lbrace{x_n}\rbrace_{n \ge m}$ is a subsequence that is a constant sequence (thus converges). Suppose $A$ is an infinite set. By $(4)$, $A$ has a limit point. Let $p$ be one limit point of $A$. By $(5)$ in the lemma, there is a sequence of points from $A$ converging to $p$. This sequence would be a subsequence of $\lbrace{x_n}\rbrace$.

$(5) \rightarrow (6)$
By the lemma, any Cauchy sequence of real numbers is bounded. Let $\lbrace{x_n}\rbrace$ be a Cauchy sequence of real numbers. By $(5)$, there is a subsequence $\lbrace{x_{n(j)}}\rbrace$ that converges. Let $p$ be the limit of the subsequence. Note that $p$ is also the limit of the original sequence $\lbrace{x_n}\rbrace$.

$(6) \rightarrow (3)$
Let $[a_n,b_n]$ be a sequence of intervals satisfying the nested interval property. Then $\lbrace{a_n}\rbrace$ is a Cauchy sequence, which converges, say to the point $p$. Then $p$ is the lone point that belongs to each $[a_n,b_n]$.

$(3) \rightarrow (1)$
Suppose that $A \subset \mathbb{R}$ is bounded above. Let $B$ be the set of all upper bounds of $A$. Note that for each $x \in A$, $x$ is a lower bound of $B$. Pick $x \in A$ and $y \in B$. Let $a_1=x$ and $b_1=y$. If there are only finitely many point of $B$ in between $a_1$ and $b_1$, then the smallest points of $B$ would be the least upper bound of $A$ and we are done. So assume that $[a_1,b_1]$ contains infinitely many points of $B$ (i.e. infinitely many upper bounds of $A$).

Consider $[a_1,m]$ and $[m,b_1]$ where $m=0.5(a_1+b_1)$. If the left interval $[a_1,m]$ contains upper bounds of $A$, we assume that that it has infinitely many of them (otherwise, the minimum would be the least upper bound and we are done). Choose one of these intervals that contains infinitely many points of $B$. If both intervals contain infinitely many points of $B$, we choose the leftmost one. Then we let $a_2$ be the left endpoint of the chosen interval and $b_2$ be the right endpoint of the interval. We observe that to the left of $a_2$, there are no upper bounds of $A$ (points of $B$).

Continue this process inductively, making sure that in each step we divide the previous interval $[a_n,b_n]$ into two equal halves. If the left interval contains upper bounds of $A$, we assume that it has infinitely many of them (otherwise, we stop at this inductive step). We choose the first half interval that contains infinitely many upper bounds of $A$ and denote it by $[a_{n+1},b_{n+1}]$. Observe that to the left of $a_{n+1}$, there are no upper bounds of $A$.

As indicated in the induction step, if we can stop at any one of the step, we would have a least upper bound of $A$ and we are done. Otherwise, we have a nested sequence of intervals $[a_n,b_n]$ such that the widths of the intervals $\rightarrow 0$. Then the intervals have exactly one point $p$ in the intersection.

We claim that $p$ is the desired least upper bound of $A$. First of all, there is no upper bounds of $A$ to the left of $p$. This stems from the observation that no upper bounds of $A$ can be found at the left side of each $a_n$ and the fact that $\left\{a_n\right\}$ converges to $p$. Thus if $p$ is an upper bound of $A$, then $p$ is the least upper bound. Note that $p$ is a limit point of $B$ since every interval $[a_n,b_n]$ contains infinitely many points of $B$. If $p$ is not an upper bound of $A$, then there is some point $q$ of $A$ such that $p. Then the interval $(p, q)$ contains no points of $B$ (upper bounds of $A$). This constradicts the fact that $p$ is a limit point of $B$. So $p$ must be an upper bound of $A$. This establishes the least upper bound property.

Reference

1. Rudin, W., Principles of Mathematical Analysis, Third Edition, 1976, McGraw-Hill, New York.