Subsection 1.2.1 The set of real numbers
We finally get to the real number system. To simplify matters, instead of constructing the real number set from the rational numbers, we simply state their existence as a theorem without proof. Notice that
\(\Q\) is an ordered field.
Theorem 1.2.1.
Note also that
\(\N \subset \Q\text{.}\) We saw that
\(1 > 0\text{.}\) By
induction (exercise), we can prove that
\(n > 0\) for all
\(n \in \N\text{.}\) Similarly, we verify simple statements about rational numbers. For example, we proved that if
\(n > 0\text{,}\) then
\(\nicefrac{1}{n} > 0\text{.}\) Then
\(m < k\) implies
\(\nicefrac{m}{n} < \nicefrac{k}{n}\text{.}\)
Analysis consists of proving inequalities, and the following proposition, or one of its many variations, is how an analyst proves a nonstrict inequality.
Proposition 1.2.2.
If
\(x \in \R\) is such that
\(x \leq \epsilon\) for all
\(\epsilon \in \R\) where
\(\epsilon > 0\text{,}\) then
\(x \leq 0\text{.}\)
Proof.
If
\(x > 0\text{,}\) then
\(0 < \nicefrac{x}{2} < x\) (why?). Take
\(\epsilon = \nicefrac{x}{2}\) to get a contradiction. Thus
\(x \leq 0\text{.}\)
For nonnegative
\(x\text{,}\) equality results:
If \(x \geq 0\) is such that \(x \leq \epsilon\) for all \(\epsilon > 0\text{,}\) then \(x = 0\text{.}\) A common version uses the absolute value (see
Section 1.3):
If \(\sabs{x} \leq \epsilon\) for all \(\epsilon > 0\text{,}\) then \(x = 0\text{.}\) To prove
\(x \geq 0\text{,}\) an analyst might prove that
\(x \geq -\epsilon\) for all
\(\epsilon > 0\text{.}\) From now on, when we say
\(x \geq 0\) or
\(\epsilon > 0\text{,}\) we automatically mean that
\(x \in \R\) and
\(\epsilon \in \R\text{.}\)
The idea behind the proposition above is that any time we have two real numbers
\(a < b\text{,}\) there is another real number
\(c\) such that
\(a < c < b\text{.}\) Infinitely many such
\(c\) exist. One of them is, for example,
\(c = \frac{a+b}{2}\) (why?). We will use this fact in the next example.
The most useful property of
\(\R\) for analysts is not just that it is an ordered field, but that it has the
least-upper-bound property. Essentially, we want
\(\Q\text{,}\) but we also want to take suprema (and infima) willy-nilly. So what we do is take
\(\Q\) and throw in enough numbers to obtain
\(\R\text{.}\)
We mentioned already that
\(\R\) contains elements that are not in
\(\Q\) because of the
least-upper-bound property. Let us prove it. We saw there is no rational square root of two. The set
\(\{ x \in \Q : x^2 < 2 \}\) implies the existence of the real number
\(\sqrt{2}\text{,}\) although this fact requires a bit of work. See also
Exercise 1.2.14.
Example 1.2.3.
Claim:
There exists a unique positive \(r \in \R\) such that \(r^2 = 2\text{.}\) We denote \(r\) by \(\sqrt{2}\text{.}\)
Proof.
Take the set
\(A \coloneqq \{ x \in \R : x^2 < 2 \}\text{.}\) We first show that
\(A\) is bounded above and nonempty. The inequality
\(x \geq 2\) implies
\(x^2 \geq 4\) (see
Exercise 1.1.3). So if
\(x^2 < 2\text{,}\) then
\(x < 2\text{.}\) So
\(A\) is bounded above. As
\(1 \in A\text{,}\) the set
\(A\) is nonempty. The supremum, therefore, exists.
Let
\(r \coloneqq \sup\, A\text{.}\) We will show that
\(r^2 = 2\) by showing that
\(r^2 \geq 2\) and
\(r^2 \leq 2\text{.}\) This is the way analysts show equality, by showing two inequalities. We already know that
\(r \geq 1 > 0\text{.}\)
In the following, it may seem we are pulling certain expressions out of a hat. When writing a proof such as this, we would, of course, come up with the expressions only after playing around with what we wish to prove. The order in which we write the proof is not necessarily the order in which we come up with the proof.
Let us first show that \(r^2 \geq 2\text{.}\) Take a positive number \(s\) such that \(s^2 < 2\text{.}\) We wish to find an \(h > 0\) such that \({(s+h)}^2 < 2\text{.}\) As \(2-s^2 > 0\text{,}\) we have \(\frac{2-s^2}{2s+1} > 0\text{.}\) Choose an \(h \in \R\) such that \(0 < h < \frac{2-s^2}{2s+1}\text{.}\) Furthermore, assume \(h < 1\text{.}\) Estimate,
\begin{equation}
\begin{aligned}
{(s+h)}^2 - s^2 & = h(2s + h) \\
& < h(2s+1) & & \quad \bigl(\text{since } h < 1\bigr) \\
& < 2-s^2 & & \quad \bigl(\text{since } h < \tfrac{2-s^2}{2s+1} \bigr) .
\end{aligned}
\end{equation}
Therefore, \({(s+h)}^2 < 2\text{.}\) Hence \(s+h \in A\text{,}\) but as \(h > 0\text{,}\) we have \(s+h > s\text{.}\) So \(s < r = \sup\, A\text{.}\) As \(s\) was an arbitrary positive number such that \(s^2 < 2\text{,}\) it follows that \(r^2 \geq 2\text{.}\)
Now take a positive number \(s\) such that \(s^2 > 2\text{.}\) We wish to find an \(h > 0\) such that \({(s-h)}^2 > 2\) and \(s-h\) is still positive. As \(s^2-2 > 0\text{,}\) we have \(\frac{s^2-2}{2s} > 0\text{.}\) Let \(h \coloneqq \frac{s^2-2}{2s}\text{,}\) and check \(s-h=s-\frac{s^2-2}{2s} = \frac{s}{2}+\frac{1}{s} > 0\text{.}\) Estimate,
\begin{equation}
\begin{aligned}
s^2 - {(s-h)}^2 & = 2sh - h^2 \\
& < 2sh & & \quad \bigl( \text{since } h^2 > 0 \text{ as } h \neq 0 \bigr) \\
& = s^2-2 & & \quad \bigl( \text{since } h = \tfrac{s^2-2}{2s} \bigr) .
\end{aligned}
\end{equation}
By subtracting \(s^2\) from both sides and multiplying by \(-1\text{,}\) we find \({(s-h)}^2 > 2\text{.}\) Therefore, \(s-h \notin A\text{.}\)
Moreover, if
\(x \geq s-h\text{,}\) then
\(x^2 \geq {(s-h)}^2 > 2\) (as
\(x > 0\) and
\(s-h > 0\)) and so
\(x \notin A\text{.}\) Thus,
\(s-h\) is an upper bound for
\(A\text{.}\) However,
\(s-h < s\text{,}\) or in other words,
\(s > r = \sup\, A\text{.}\) Hence,
\(r^2 \leq 2\text{.}\)
Together,
\(r^2 \geq 2\) and
\(r^2 \leq 2\) imply
\(r^2 = 2\text{.}\) The existence part is finished. We still need to handle uniqueness. Suppose
\(s \in \R\) is such that
\(s^2 = 2\) and
\(s > 0\text{.}\) Thus,
\(s^2 = r^2\text{.}\) However, if
\(0 < s < r\text{,}\) then
\(s^2 < r^2\text{.}\) Similarly,
\(0 < r < s\) implies
\(r^2 < s^2\text{.}\) Hence,
\(s = r\text{.}\)
The number
\(\sqrt{2} \notin \Q\text{.}\) The set
\(\R \setminus \Q\) is called the set of
irrational numbers. We just proved that
\(\R \setminus \Q\) is nonempty. Not only is it nonempty, as we will see, it is very large indeed.
Using the same technique as above, we can show that a positive real number
\(x^{1/n}\) exists for all
\(n\in \N\) and all
\(x > 0\text{.}\) That is, for each
\(x > 0\text{,}\) there exists a unique positive real number
\(r\) such that
\(r^n = x\text{.}\) The proof is left as an exercise.
Subsection 1.2.3 Using supremum and infimum
Suprema and infima are compatible with algebraic operations. For a set \(A \subset \R\) and \(x \in \R\) define
\begin{equation}
\begin{aligned}
x + A & \coloneqq \{ x+y \in \R : y \in A \} , \\
xA & \coloneqq \{ xy \in \R : y \in A \} .
\end{aligned}
\end{equation}
For example, if \(A = \{ 1,2,3 \}\text{,}\) then \(5+A = \{ 6,7,8 \}\) and \(3A = \{ 3,6,9
\}\text{.}\)
Proposition 1.2.6.
Let
\(A \subset \R\) be nonempty.
-
If
\(x \in \R\) and
\(A\) is bounded above, then
\(\sup (x+A) = x + \sup\, A\text{.}\)
-
If
\(x \in \R\) and
\(A\) is bounded below, then
\(\inf (x+A) = x + \inf\, A\text{.}\)
-
If
\(x > 0\) and
\(A\) is bounded above, then
\(\sup (xA) = x ( \sup\, A )\text{.}\)
-
If
\(x > 0\) and
\(A\) is bounded below, then
\(\inf (xA) = x ( \inf\, A )\text{.}\)
-
If
\(x < 0\) and
\(A\) is bounded below, then
\(\sup (xA) = x ( \inf\, A )\text{.}\)
-
If
\(x < 0\) and
\(A\) is bounded above, then
\(\inf (xA) = x ( \sup\, A )\text{.}\)
Note how multiplying a set by a negative number swaps suprema and infima. Also, as the proposition implies that the supremum (resp. infimum) of
\(x+A\) or
\(xA\) exists, it also implies that
\(x+A\) or
\(xA\) is nonempty and bounded above (resp. below).
Proof.
Let us only prove the first statement. The rest are left as exercises.
Suppose \(b\) is an upper bound for \(A\text{.}\) That is, \(y \leq b\) for all \(y \in A\text{.}\) Then \(x+y \leq x+b\) for all \(y \in A\text{,}\) and so \(x+b\) is an upper bound for \(x+A\text{.}\) In particular, if \(b = \sup\, A\text{,}\) then
\begin{equation}
\sup (x+A) \leq x+b = x+ \sup\, A .
\end{equation}
The opposite inequality is similar. If \(c\) is an upper bound for \(x+A\text{,}\) then \(x+y \leq c\) for all \(y \in A\) and so \(y \leq c-x\) for all \(y \in A\text{.}\) So \(c-x\) is an upper bound for \(A\text{.}\) If \(c = \sup (x+A)\text{,}\) then
\begin{equation}
\sup\, A \leq c-x = \sup (x+A) -x .
\end{equation}
The result follows.
Sometimes we need to apply supremum or infimum twice. Here is an example.
Proposition 1.2.7.
Let
\(A, B \subset \R\) be nonempty sets such that
\(x \leq y\) whenever
\(x \in A\) and
\(y \in B\text{.}\) Then
\(A\) is bounded above,
\(B\) is bounded below, and
\(\sup\, A \leq \inf\, B\text{.}\)
Proof.
Any
\(x \in A\) is a lower bound for
\(B\text{.}\) Therefore,
\(x \leq \inf\, B\) for all
\(x \in A\text{,}\) so
\(\inf\, B\) is an upper bound for
\(A\text{.}\) Hence,
\(\sup\, A \leq \inf\, B\text{.}\)
We must be careful about strict inequalities and taking suprema and infima. Note that
\(x < y\) whenever
\(x \in A\) and
\(y \in B\) still only implies
\(\sup\, A \leq \inf\, B\text{,}\) and not a strict inequality. For example, take
\(A \coloneqq \{ 0 \}\) and
\(B \coloneqq \{ \nicefrac{1}{n}
: n \in \N \}\text{.}\) Then
\(0 < \nicefrac{1}{n}\) for all
\(n \in \N\text{.}\) However,
\(\sup\, A = 0\) and
\(\inf\, B = 0\text{.}\) This important subtle point comes up often.
The proof of the following often-used fact is left to the reader. A similar result holds for infima.
Proposition 1.2.8.
If
\(S \subset \R\) is nonempty and bounded above, then for every
\(\epsilon > 0\) there exists an
\(x \in S\) such that
\((\sup\, S) - \epsilon < x \leq \sup\, S\text{.}\)
To make using suprema and infima even easier, we may want to write
\(\sup\, A\) and
\(\inf\, A\) without worrying about
\(A\) being bounded and nonempty. We make the following natural definitions.
Definition 1.2.9.
Let
\(A \subset \R\) be a set.
-
If
\(A\) is empty, then
\(\sup\, A \coloneqq -\infty\text{.}\)
-
If
\(A\) is not bounded above, then
\(\sup\, A \coloneqq \infty\text{.}\)
-
If
\(A\) is empty, then
\(\inf\, A \coloneqq \infty\text{.}\)
-
If
\(A\) is not bounded below, then
\(\inf\, A \coloneqq -\infty\text{.}\)
The statements in the definition above follow naturally if, instead of working in the ordered set \(\R\text{,}\) we take suprema and infima in the ordered set \(\R^* \coloneqq \R \cup \{ -\infty , \infty\}\) , where we let
\begin{equation}
-\infty < \infty \quad \text{and} \quad
-\infty < x \quad \text{and} \quad
x < \infty \quad \text{for all $x \in \R$}.
\end{equation}
The set \(\R^*\) is called the set of extended real numbers. For convenience, \(\infty\) and \(-\infty\) are sometimes treated as if they were numbers, except we do not allow arbitrary arithmetic with them. It is possible to define some arithmetic on \(\R^*\text{.}\) Most operations are extended in an obvious way, but we must leave \(\infty-\infty\text{,}\) \(0 \cdot (\pm\infty)\text{,}\) and \(\frac{\pm\infty}{\pm\infty}\) undefined. We refrain from using this arithmetic; it leads to easy mistakes as \(\R^*\) is not a field. Now we can take suprema and infima without fear of emptiness or unboundedness. In this book, we mostly avoid using \(\R^*\) outside of exercises and leave such generalizations to the interested reader.
