Axiom 1.2.1. The Completeness Axiom of \(\R\).
Every non-empty set of real numbers that is bounded above has a supremum in \(\R\text{.}\)
Section 1.2 The Completeness Axiom
The following axiom is what sets the real field apart from other ordered fields.
For an ordered set, the least upper bound property means that every nonempty subset that is bounded above has a least upper bound, that is, a supremum, in the set. Thus, the completeness axiom says that \(\R\text{,}\) with its usual order, has the least upper bound property. Dually, this is equivalent to saying that every nonempty subset of \(\R\) that is bounded below has an infimum in \(\R\text{;}\) see Exercise 1.3.9.
Subsection 1.2.1 Consequences
An ordered field is Archimedean if for every \(x\) in the field there exists \(n \in \N\) such that \(x < n\text{.}\)
Proposition 1.2.2.
The ordered field \(\R\) is Archimedean.
Proof.
By the definition above, it suffices to show that for every \(x \in \R\) there exists \(n \in \N\) such that \(x < n\text{.}\)
Suppose instead that some \(x \in \R\) were an upper bound for \(\N\text{.}\) Then \(\N\) would be non-empty and bounded above, so by the least upper bound property, \(u:=\sup\N\) would exist. Since \(u-1<u\text{,}\) the number \(u-1\) cannot be an upper bound of \(\N\text{.}\) So there is \(n\in\N\) with \(u-1<n\text{.}\) Then \(u<n+1\text{,}\) and \(n+1\in\N\text{,}\) contradicting that \(u\) is an upper bound of \(\N\text{.}\)
Therefore \(\N\) is not bounded above in \(\R\text{,}\) and hence \(\R\) is Archimedean.
Proposition 1.2.3.
For any \(x \in \R\) and any \(\varepsilon > 0\text{,}\) there exists \(N \in \N\) such that \(N\varepsilon > x\text{.}\)
Proof.
Since \(\varepsilon > 0\text{,}\) the number \(x/\varepsilon\) is real. By Proposition 1.2.2, there exists \(N \in \N\) such that \(N > x/\varepsilon\text{.}\) Multiplying by \(\varepsilon\) gives \(N\varepsilon > x\text{.}\)
Proposition 1.2.4.
For any \(\varepsilon > 0\text{,}\) there exists \(n \in \N\) such that \(1/n < \varepsilon\text{.}\)
Proof.
Apply Proposition 1.2.3 with \(x=1\text{.}\) Then there exists \(n \in \N\) such that \(n\varepsilon > 1\text{.}\) Since \(n>0\text{,}\) dividing by \(n\varepsilon\) yields \(1/n < \varepsilon\text{.}\)
Proposition 1.2.5.
For any \(x \in \R\text{,}\) there exists \(n \in \Z\) such that \(n \le x < n+1\text{.}\)
Proof.
By Proposition 1.2.2, there exists \(N \in \N\) such that \(N > |x|\text{.}\) Then \(-N < x < N\text{.}\)
Consider the finite set
\begin{equation*}
S=\{-N,-N+1,\dots,0,\dots,N\}\subseteq\Z.
\end{equation*}
Since \(-N \le x\text{,}\) the set of integers in \(S\) that are at most \(x\) is non-empty. Let \(n\) be the largest such integer.
Then \(n \le x\text{.}\) Also \(n \ne N\text{,}\) because \(x < N\text{.}\) Hence \(n+1 \in S\text{.}\) By maximality of \(n\text{,}\) the integer \(n+1\) cannot satisfy \(n+1 \le x\text{.}\) Therefore \(x < n+1\text{,}\) and so \(n \le x < n+1\text{.}\)
Proposition 1.2.6.
For any \(x,y \in \R\text{,}\) if \(y-x > 1\text{,}\) then there exists \(m \in \Z\) such that \(x \le m < y\text{.}\)
Proof.
If \(y \le n+1\text{,}\) then
\begin{equation*}
n \le x < y \le n+1,
\end{equation*}
so \(y-x < (n+1)-n = 1\text{,}\) contradicting the hypothesis. Therefore \(n+1 < y\text{.}\)
Let \(m=n+1\text{.}\) Then \(m \in \Z\) and \(x < m < y\text{.}\) In particular, \(x \le m < y\text{.}\)
A subset \(D\) of \(\R\) is dense if between any two real numbers lies an element of \(D\text{.}\)
Proposition 1.2.7.
The set \(\Q\) is dense in \(\R\text{.}\)
Proof.
Let \(x,y \in \R\) with \(x<y\text{.}\) Then \(y-x>0\text{,}\) so by Proposition 1.2.3 there exists \(N \in \N\) such that \(N(y-x)>1\text{.}\)
Applying Proposition 1.2.6 to the numbers \(Nx\) and \(Ny\text{,}\) we obtain an integer \(m\) such that \(Nx < m < Ny\text{.}\)
Since \(N>0\text{,}\) dividing by \(N\) gives \(x < m/N < y\text{.}\) Because \(m/N \in \Q\text{,}\) this proves that \(\Q\) is dense in \(\R\text{.}\)
Theorem 1.2.8. Existence of Positive Square Roots.
For each \(a>0\text{,}\) there exists a unique \(u>0\) such that \(u^2=a\text{.}\)
Proof.
Uniqueness is immediate because \(x\mapsto x^2\) is strictly increasing on \((0,\infty)\text{.}\)
For existence, define \(S=\{x\in\R : x>0,\ x^2\le a\}\text{.}\) The set is non-empty: if \(a\ge 1\text{,}\) then \(1\in S\text{;}\) if \(0<a<1\text{,}\) then \(a\in S\text{.}\) It is bounded above by \(M=\max\{a,1\}\text{,}\) since \(x>M\) implies \(x^2>a\text{.}\) By completeness, \(u:=\sup S\) exists.
If \(u^2<a\text{,}\) then for sufficiently small \(\varepsilon>0\text{,}\) we still have \((u+\varepsilon)^2\le a\text{,}\) so \(u+\varepsilon\in S\text{,}\) contradicting that \(u\) is an upper bound of \(S\text{.}\) If \(u^2>a\text{,}\) then for sufficiently small \(\varepsilon>0\text{,}\) \((u-\varepsilon)^2>a\text{;}\) then every \(x\in S\) satisfies \(x\le u-\varepsilon\text{,}\) so \(u-\varepsilon\) is an upper bound of \(S\text{,}\) contradicting leastness of \(u\text{.}\) Hence \(u^2=a\text{.}\)
A standard comparison example is the ordered field \(\Q\text{:}\) the set \(\{q\in\Q:q>0,\ q^2<2\}\) is non-empty and bounded above in \(\Q\text{,}\) but it has no supremum in \(\Q\text{.}\) Therefore \(\Q\) does not satisfy the least upper bound property.
