Published Jun 18, 2021
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The first application of the Axiom of Completeness is a result that may look like a more natural way to mathematically express the sentiment that the real line contains no gap.
Theorem 1.4.1 Nested Interval Property For each \(n \in N\), assume we are given a closed interval \(I_n = [a_n, b_n] = \{x \in R : a_n \leqslant x \leqslant b_n\}\). Assume also that each \(I_n\) contains \(I_{n + 1}\). Then, the resulting nested sequence of closed intervals $$ I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \cdot \cdot \cdot $$ has a nonempty intersection; that is, \(\bigcap_{n=1}^\infty I_n \neq \emptyset\)
Proof. In order to show that \(\bigcap_{n=1}^\infty I_n\) is not empty, we are going to use the Axiom of Completeness (AoC) to produce a single real number \(x\) satisfying \(x \in I_n\) for every \(n \in N\). Now, Aoc is a statement about bounded sets, and the one we want to consider is the set $$ A = \{a_n : n \in N\} $$ of left-hand endpoints of the intervals.
Because the intervals are nested, we see that every \(b_n\) servers as an upper bound for \(A\). Thus, we are justified in setting $$ x = sup A. $$ Now, consider a particular \(I_n = [a_n, b_n]\). Because \(x\) is an upper bound for \(A\), we have \(a_n \leqslant x\). The fact that each \(b_n\) is an upper bound for \(A\) and the \(x\) is the least upper bound implies \(x \leqslant b_n\).
Altogether then, we have \(a_n \leqslant x \leqslant b_n\), which means \(x \in I_n\) for every choice of \(n \in N\). Hence, \(\bigcap_{n=1}^\infty I_n\), and the intersection is not empty.
The set \(Q\) is an extension of \(N\), and \(R\) in turn is an extension of \(Q\). The next few results indicate how \(N\) and \(Q\) sit inside of \(R\).
Theorem 1.4.2 Archimedean Property
Proof. Part \((i)\) of the proposition states that \(N\) is not bounded above. There has never been any doubt about the truth of this, and it could be reasonably argued that we should not have to prove it at all, especially in light of the fact that we have decided to take other familiar properties of \(N\), \(Z\), and \(Q\) as given.
The counterargument is that there is still a great deal of mystery about what real numbers actually are. What we have said so far is that \(R\) is an extension of \(Q\) that maintains the algebraic and order properties of the rationals but also possesses the least upper bound property articulated in the Axiom of Completeness. In the absence of any other information about \(R\), we have to consider the possibility that in extending \(Q\) we unwittingly acquired some new numbers that are upper bounds for \(N\). In fact, as disorienting as it may sound, there are ordered filed extensions of \(Q\) that include "numbers" bigger than every natural number. It asserts that the real numbers do not contain such exotic creatures. The Axiom of Completeness, which we adopted to patch the holes in \(Q\), carries with it the implication that \(N\) is an unbounded subset of \(R\).
And so to the proof. Assume, for contradiction, that \(N\) is bounded above. By the Axiom of Completeness (AoC), \(N\) should then have a least upper bound, and we can set \(\alpha = sup N\). If we consider \(\alpha - 1\), then we no longer have an upper bound, and therefore there exists an \(n \in N\) satisfying (\\alpha - 1 < n\). But this is equivalent to \(\alpha < n + 1\). Because \(n + 1 \in N\), we have a contraction to the fact that \(\alpha\) is supposed to be an upper bound for \(N\). (Notice that the contradiction here depends only on AoC and the fact the \(N\) is closed under addition)
Part \((ii)\) follows from \((i)\) by letting \(x = 1/y\)
This familiar property of \(N\) is the key to an extremely important fact about how \(Q\) fits inside \(R\).
Theorem 1.4.3 (Density of \(Q\) in \(R\)). For every two real numbers \(a\) and \(b\) with \(a < b\), there exists a rational number \(r\) satisfying \(a < r < b\)
Proof. A rational number is q quotient of integers, so we must produce \(m \in Z\) and \(n \in N\) so That $$ a < \frac{m}{n} < b \tag{1}\label{1} $$ The first step is to choose the denominator \(n\) large enough so that consecutive increments of size \(1/n\) are too close together to "step over" the interval \((a, b)\)
Using the Archimedean Property, we may pick \(n \in N\) large enough so that $$ \frac{1}{n} < b - a \tag{2}\label{2} $$ Inequality \(\eqref{1}\) is equivalent to \(na < m < nb\). With \(n\) already chosen, the idea now is to choose \(m\) to be the smallest integer greater than \(na\). In other words, pick \(m \in Z\) so that $$ m - 1 \leqslant na < m \tag{3}\label{3} $$ Now, inequality \(\eqref{3}\) immediately yields \(a < m/n\), which is half of the battle. Inequality \(\eqref{2}\) is equivalent to \(a < b - 1/n\),we can use \(\eqref{3}\) to write $$ \begin{align} m & \leqslant na + 1 \\ & < n(b - \frac{1}{n}) + 1 \\ & = nb \end{align} $$ Because \(m < nb\) implies \(m/n < b\), we have \(a < m/n < b\), as desired.
Corollary 1.4.4 Given any two real numbers \(a < b\), there exists an irrational number \(t\) satisfying \(a < t < b\)
Theorem 1.4.5 There exists a real number \(a \in R\) satisfying \(\alpha^2 = 2\)
Proof. Consider the set $$ T = \{t \in R : t^2 < 2\} $$ and set \(\alpha = sup T\). We are going to prove \(\alpha^2 = 2\) by ruling out the possibilities \(\alpha^2 < 2\) and \(\alpha^2 > 2\). Keep in mind that there are two parts to the definition of \(sup T\), and they will both be important. (This always happens when a supremum is used in an argument.) The strategy is to demonstrate that \(\alpha^2 < 2\) violates the fact that the \(\alpha\) is an upper bound for \(T\), and \(\alpha^2 > 2\) violates the fact that it is the least upper bound.
Let's first see what happens if we assume \(\alpha^2 < 2\). In search of an element of \(T\) that is larger than \(\alpha\), write $$ \begin{align} (\alpha + \frac{1}{n})^2 &= \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n^2} \\ &< \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n} \\ &= \alpha^2 + \frac{2\alpha + 1}{n} \end{align} $$ But assuming \(\alpha^2 < 2\) gives us a little space in which to fit the \((2\alpha + 1)/n\) term and keep the total less than \(2\). Specifically, chose \(n_0 \in N\) large enough so that $$ \frac{1}{n_0} < \frac{2 - \alpha^2}{2\alpha + 1} $$ This implies \((2\alpha + 1)/n_0 < 2 - \alpha^2\), and consequently that $$ (\alpha + \frac{1}{n_0})^2 < \alpha^2 + (2 - \alpha^2) = 2 $$ Thus, \(\alpha + 1/n_0 \in T\), contradicting the fact that \(\alpha\) is an upper bound for \(T\). We conclude that \(\alpha^2 < 2\) cannot happen.
Now, what about the case \(\alpha^2 > 2\)? We write $$ \begin{align} (\alpha - \frac{1}{n})^2 &= \alpha^2 - \frac{2\alpha}{n} + \frac{1}{n^2} \\ &> \alpha^2 - \frac{2\alpha}{n} \end{align} $$
A small modification of this proof can be made to show that \(\sqrt{x}\) exists for any \(x \leqslant 0\). A formula for expanding \((\alpha + 1/n)^m\) called the binomial formula can be used to show that \(\sqrt[m]{x}\) exists for arbitrary values of \(m \in N\)