Continuity of real numbers

We know that $\mathbb{R}$ (the set of real numbers) and $\mathbb{Q}$ (the set of rational numbers) share similar properties. Both are fields and dense. What distinguishes real numbers from rational numbers is continuity. We prove Archimedes' principle as a consequence of the continuity of real numbers.

Definition (Upper bound, lower bound)

Let $S\subset \mathbb{R}$ and $\alpha \in \mathbb{R}$.

1.  If, for all $x\in S$, $x\le \alpha$, then $\alpha$ is said to be an upper bound of $S$.
2. If, for all $x\in S$, $x\ge \alpha$, then $\alpha$ is said to be a lower bound of $S$.

Example. Let $A=\{x\mid x\le 1,x\in \mathbb{R}\}$. Then, 2 is an upper bound of $A$. 1.5 is another upper bound of $A$. In fact, any number greater than or equal to 1 is an upper bound of $A$. There are no lower bounds of $A$. □

Example. Let $B=\{x\mid x>\sqrt{2},x\in \mathbb{Q}\}$. Any number less than or equal to $\sqrt{2}$ is a lower bound of $B$. There are no upper bounds of $B$. □

The following lemma should be trivial.

Lemma

Let $S\subset \mathbb{R}$ and $a\in \mathbb{R}$.

1. $a$ is not an upper bound of $S$ if and only if $\mathrm{\exists }x\in S(x>a)$.
2. $a$ is not a lower bound of $S$ if and only if $\mathrm{\exists }x\in S(x<a)$.

Definition (Bounded above/below)

Let $S\subset \mathbb{R}$. If $S$ has an upper bound, then $S$ is said to be bounded above. If $S$ has a lower bound, then $S$ is said to be bounded below. If $S$ is bounded above and below, then $S$ is said to be bounded.

The following lemma should be also trivial.

Lemma

Let $S\subset \mathbb{R}$ and define

$$\begin{array}{rcl}U(S)& =& \{a\mid \mathrm{\forall }x\in S(x\le a)\},\\ L(S)& =& \{a\mid \mathrm{\forall }x\in S(x\ge a)\}.\end{array}$$

That is, $U(S)$ is the set of upper bounds of $S$, and $L(S)$ is the set of lower bounds of $S$. Then

1. $S$ is bounded above if and only if $U(S)\ne \mathrm{\varnothing }$.
2. $S$ is bounded below if and only if $L(S)\ne \mathrm{\varnothing }$.

Definition (Supremum and infimum)

Let $S\subset \mathbb{R}$.

1. If $S$ is bounded above, the least upper bound, if it exists, is called the supremum of $S$ and is denoted $supS$.
2. If $S$ is bounded below, the greatest lower bound, if it exists, is called the infimum of $S$ and is denoted $infS$.

In other words, $supS=minU(S)$ and $infS=maxL(S)$. 

Remark. It may be that $supS\in S$ or $supS\notin S$. Same for $infS$. □

Example. 

1. If $A=\{x\mid x\le 1,x\in \mathbb{R}\}$, then $supA=1\in A$. 
2. If $B=\{x\mid x\ge \sqrt{2},x\in \mathbb{Q}\}$, then $infB=\sqrt{2}\notin B$. (Note that $B$ is a subset of $\mathbb{Q}$.)

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In the closed interval $[a,b]=\{x\mid a\le x\le b,x\in \mathbb{R}\}$, $b$ is the maximum value which is also the supremum. Similarly, $a$ is the minimum value which is also the infimum. In general,

Lemma

1. If $S$ has the maximum $b$, then $b=supS$.
2. If $S$ has the minimum $a$, then $a=infS$.

On the other hand, the open interval $(a,b)=\{x\mid a<x<b,x\in \mathbb{R}\}$ has neither maximum nor minimum. Yet we can still define its supremum and infimum (namely, $b$ and $a$, respectively). Therefore, in a sense, supremum and infimum are a generalization of maximum and minimum.

We may redefine supremum and infimum in the following manner.

Definition (Supremum and infimum (second take))

Let $S\subset \mathbb{R}$.

1. $a\in \mathbb{R}$ is the supremum of $S$ if the following two conditions are met:
1. $\mathrm{\forall }x\in S(x\le a)$. (i.e., $a$ is an upper bound.)
2. If $a^{\prime }<a$, then $\mathrm{\exists }x\in S(a^{\prime }<x)$. (i.e., $a$ is the least upper bound.)
2. $a\in \mathbb{R}$ is the infimum of $S$ if the following two conditions are met.
1. $\mathrm{\forall }x\in S(x\ge a)$. (i.e., $a$ is a lower bound.)
2. If $a^{\prime }>a$, then $\mathrm{\exists }x\in S(a^{\prime }>x)$. (i.e., $a$ is the greatest lower bound.)

Lemma

Let $S,T$ be subsets of $\mathbb{R}$ such that $T\subset S$. Then

1. $supS\ge supT$.
2. $infS\le infT$.

Proof. In the following, $U(S)$ and $L(S)$ denote the set of upper bounds of $S$ and the set of lower bounds of $S$, respectively.

1. Suppose $a\in U(S)$. Then for all $x\in S$, we have $x\le a$. However, since $T\subset S$, for all $x\in T$, we have $x\le a$. In short, $a\in U(S)⟹a\in U(T)$. Hence $U(S)\subset U(T)$. Therefore, the minimum of $U(S)$ is greater than or equal to the minimum of $U(T)$. Thus $supS\ge supT$. (You should examine why $U(S)\subset U(T)$ implies $minU(S)\ge minU(T)$.)
2. Similarly, we have $L(S)\subset L(T)$. Hence $infS\le infT$.

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$\mathbb{R}\setminus \mathbb{Q}$ is the set of irrational numbers. Thus, if we consider $\mathbb{Q}$ in $\mathbb{R}$, there are many ``spaces'' filled by irrational numbers. But $\mathbb{R}$ has no unfilled spaces. In this sense, $\mathbb{R}$ is ``continuous.'' To see what this exactly means, consider the following example.

Example. $S=\{x\mid x^2<2,x\in \mathbb{R}\}$ is a subset of $\mathbb{R}$ with $supS=\sqrt{2}$ and $infS=-\sqrt{2}$. And, of course, $\pm \sqrt{2}\in \mathbb{R}$, that is, both the supremum and infimum are members of $\mathbb{R}$.

  In contrast, if we only consider $\mathbb{Q}$, then the supremum and infimum do not exist within $\mathbb{Q}$. □

So, the very existence of supremum and infimum within $\mathbb{R}$ illuminates the ``continuity'' of real numbers. Based on this observation, we adopt the following axioms as fundamental properties of real numbers.

Axiom (Continuity of real numbers)

Let $S\subset \mathbb{R}$.

1. If $S$ is bounded above, then $supS\in \mathbb{R}$.
2. If $S$ is bounded below, then $infS\in \mathbb{R}$.

From this axiom, we can prove Archimedes' principle as a theorem.

Theorem (Archimedes' principle)

For any positive real number $a$ and any real number $b$, there exists a natural number $n$ such that $an>b$.

Remark. Recall that, previously, we adopted Archimedes' principle as an axiom. See Some well-known sets of numbers. □

Proof. We prove it by contradiction. 

Suppose that there exists a positive real number $a$ and a real number $b$ such that, for any natural number $n$,  $an\le b$.

That is, for any natural number $n$, we have $n\le \frac{b}{a}$. Therefore $\mathbb{N}$ is bounded above. By the continuity axiom, the supremum of $\mathbb{N}$ exists. Let $s=sup\mathbb{N}$. Since $s-1$ is not the supremum of $\mathbb{N}$, there exists $m\in \mathbb{N}$ such that $s-1<m$. Therefore $s<m+1$. But $m+1$ is a natural number that is greater than the supremum $s$ of $\mathbb{N}$, which is a contradiction. ■

Corollary

For any positive real number $a$ and any real number $b$, there exists a natural number $n$ such that $a>\frac{b}{n}$.

Proof. (Exercise.) ■

Example. Let $A=\{\frac{1}{n}\mid n\in \mathbb{N}\}$. Then, $infA=0$.

In fact, for any $n\in \mathbb{N}$, $0<\frac{1}{n}$ so that $0$ is a lower bound of $A$. Let $a$ be an arbitrary positive real number. By the above corollary, there exists a natural number $n$ such that $a>\frac{1}{n}$. This implies that no positive real numbers are a lower bound of $A$. Thus, all non-positive real numbers are lower bounds of $A$, and 0 is the greatest lower bound. Thus, $infA=0$. □

Example. Let $a,b\in \mathbb{R}$ be given. Suppose, for any positive real number $\epsilon$, $|a-b|<\epsilon$. Then $a=b$.

This can be proved by contradiction.

Suppose $a\ne b$. Then $|a-b|$ is a positive real number so that there exists $n\in \mathbb{N}$ such that

$$\begin{array}{}\text{(eq:abn)}& |a-b|>\frac{1}{n}.\end{array}$$

But $\frac{1}{n}$ is also a positive real number. Thus (eq:abn) contradicts the assumption that $|a-b|<\epsilon$ for any $\epsilon >0$. □

Previously, we said that $\mathbb{Q}$ had ``spaces'' only filled by irrational numbers. Yet, $\mathbb{Q}$ is ``dense''.

Theorem (Denseness of rational numbers)

A non-empty open interval contains at least one rational number.

Remark. We have already proved this in Some well-known sets of numbers. We prove it here, again, anyway. □

Proof. Consider an open interval $(a,b)$ where $a<b$. If $a<0<b$, then $(a,b)$ contains the rational number $0$. If $a<b<0$, if $(-b,-a)$ contains a rational number $r$, then $(a,b)$ contains the rational number $-r$. Therefore it suffices to prove the case for $0<a<b$. 

By Archimedes' principle, we can find a natural number $n$ such that $(b-a)n>1$. Then we have $a+\frac{1}{n}<b$. Applying Archimedes' principle again, we can find a natural number $m$ such that $\frac{1}{n}m>a$. Taking the minimum such natural number $m$, we have

$$\frac{m-1}{n}\le a<\frac{m}{n}.$$

But then we have

$$a<\frac{m}{n}=\frac{m-1}{n}+\frac{1}{n}\le a+\frac{1}{n}<b.$$

Thus, the rational number $\frac{m}{n}$ belongs to the open interval $(a,b)$.  ■

Corollary

Let $\alpha$ be any real number and $\epsilon$ be any positive real number. Then, there exists at least one rational number $a$ such that $|\alpha -a|<\epsilon$. 

Proof. Consider the open interval $(\alpha -\epsilon ,\alpha +\epsilon )$. Since $\epsilon >0$, this interval is not empty. By the above theorem, there exists a rational number $a$ in this interval. Since $a\in (\alpha -\epsilon ,\alpha +\epsilon )$, we have

$$\alpha -\epsilon <a<\alpha +\epsilon$$

so that

$$|\alpha -a|<\epsilon .$$

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This result shows that any real number can be approximated by a rational number arbitrarily accurately.

Remark. We have used this property to derive the inverse of the natural logarithm, $\exp (x)$, is equal to $e^x$ for all $x\in \mathbb{R}$ in $\log$ and $e$. □

Example. 

1. Consider $|\sqrt{2}-a|<0.01$. $a=1.41$ qualifies, $a=1.43$ does not.
2. $\pi =3.141592\cdots$ is an irrational number. Consider $|\pi -a|$. $a=3.14$ is a rational number that does not satisfy $|\pi -a|<0.001$, but it does satisfy $|\pi -a|<0.002$. 

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