Continuity of real numbers

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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 \(x\in S\), \(x \leq \alpha\), then \(\alpha\) is said to be an upper bound of \(S\).
  2. If, for all \(x\in S\), \(x \geq \alpha\), then \(\alpha\) is said to be a lower bound of \(S\).
Example. Let \(A = \{x \mid x \leq 1, x \in \mathbb{R}\}\). 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 \(\exists x\in S ~(x > a)\).
  2. \(a\) is not a lower bound of \(S\) if and only if \(\exists x \in S ~(x < a)\).

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
\[ \begin{eqnarray} U(S) &=& \{a \mid \forall x\in S ~(x\leq a)\},\\ L(S) &=& \{a \mid \forall x\in S ~(x\geq a)\}. \end{eqnarray} \]
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 \(U(S) \neq \emptyset\).
  2. \(S\) is bounded below if and only if \(L(S) \neq \emptyset\).

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 \(\sup S\).
  2. If \(S\) is bounded below, the greatest lower bound, if it exists, is called the infimum of \(S\) and is denoted \(\inf S\).
In other words, \(\sup S = \min U(S)\) and \(\inf S = \max L(S)\). 

Remark. It may be that \(\sup S \in S\) or \(\sup S \not\in S\). Same for \(\inf S\). □

Example. 
  1. If \(A = \{x \mid x \leq 1, x \in \mathbb{R}\}\), then \(\sup A = 1 \in A\). 
  2. If \(B = \{x \mid x \geq \sqrt{2}, x \in \mathbb{Q}\}\), then \(\inf B = \sqrt{2} \not\in B\). (Note that \(B\) is a subset of \(\mathbb{Q}\).)
□
In the closed interval \([a, b] = \{x \mid a \leq x \leq b, x\in\mathbb{R}\}\), \(b\) 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 \(b = \sup S\).
  2. If \(S\) has the minimum \(a\), then \(a = \inf S\).
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. \(\forall x\in S ~ (x \leq a)\). (i.e., \(a\) is an upper bound.)
    2. If \(a' < a\), then \(\exists x\in S ~ (a' < x)\). (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. \(\forall x\in S ~ (x \geq a)\). (i.e., \(a\) is a lower bound.)
    2. If \(a' > a\), then \(\exists x\in S ~ (a' > x)\). (i.e., \(a\) is the greatest lower bound.)

Lemma

Let \(S, T\) be subsets of \(\mathbb{R}\) such that \(T \subset S\). Then
  1. \(\sup S \geq \sup T\).
  2. \(\inf S \leq \inf T\).
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 \(x \leq a\). However, since \(T \subset S\), for all \(x\in T\), we have \(x \leq a\). In short, \(a\in U(S) \implies a \in U(T)\). Hence \(U(S) \subset U(T)\). Therefore, the minimum of \(U(S)\) is greater than or equal to the minimum of \(U(T)\). Thus \(\sup S \geq \sup T\). (You should examine why \(U(S)\subset U(T)\) implies \(\min U(S) \geq \min U(T)\).)
  2. Similarly, we have \(L(S) \subset L(T)\). Hence \(\inf S \leq \inf T\).
■

\(\mathbb{R}\setminus\mathbb{Q}\) 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 \(\sup S = \sqrt{2}\) and \(\inf S = -\sqrt{2}\). 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 \(\sup S \in \mathbb{R}\).
  2. If \(S\) is bounded below, then \(\inf S \in \mathbb{R}\).
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 \(n\),  \(an \leq b\).

That is, for any natural number \(n\), we have \(n \leq \frac{b}{a}\). 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 \(A = \left\{\frac{1}{n}\mid n \in\mathbb{N}\right\}\). Then, \(\inf A = 0\).
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, \(\inf A = 0\). □

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

This can be proved by contradiction.

Suppose \(a\neq b\). Then \(|a - b|\) is a positive real number so that there exists \(n\in\mathbb{N}\) such that
\[|a - b| > \frac{1}{n}.\tag{eq:abn}\]
But \(\frac{1}{n}\) is also a positive real number. Thus (eq:abn) contradicts the assumption that \(|a - b| < \varepsilon\) for any \(\varepsilon > 0\). □

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
\[\frac{m-1}{n} \leq a < \frac{m}{n}.\]
But then we have
\[a < \frac{m}{n} = \frac{m-1}{n} + \frac{1}{n} \leq a + \frac{1}{n} < b.\]
Thus, the rational number \(\frac{m}{n}\) belongs to the open interval \((a, b)\).  ■

Corollary

Let \(\alpha\) be any real number and \(\varepsilon\) be any positive real number. Then, there exists at least one rational number \(a\) such that \(|\alpha - a| < \varepsilon\). 
Proof. Consider the open interval \((\alpha - \varepsilon, \alpha + \varepsilon)\). Since \(\varepsilon > 0\), this interval is not empty. By the above theorem, there exists a rational number \(a\) in this interval. Since \(a \in (\alpha - \varepsilon, \alpha + \varepsilon)\), we have
\[\alpha - \varepsilon < a <  \alpha + \varepsilon\]
so that
\[|\alpha - a| < \varepsilon.\]
■
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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