8.10: Order-Completeness
We give an argument for the uncountability of \(\mathbb{R}\) depending only on its abstract order properties.
DEFINITION. Order-complete Let \((X, \leq)\) be a linearly ordered set. It is called order-complete if, whenever \(A\) and \(B\) are non-empty subsets of \(X\) with the property that \[(\forall a \in A)(\forall b \in B) \quad a \leq b,\] then there exists \(c\) in \(X\) such that \[(\forall a \in A)(\forall b \in B) \quad a \leq c \leq b .\] Note that any order-complete set must have the least upper bound property - if \(A\) is any non-empty bounded set, let \(B\) be the set of all upper bounds for \(A\) , and then \(c\) from (8.20) is the (unique) least upper bound for \(A\) .
DEFINITION. Dense Let \((X, \leq)\) be a linearly ordered set, and \(Y \subseteq\) \(X\) . We say \(Y\) is dense in \(X\) if \[(\forall a<b \in X)(\exists y \in Y) a<y<b .\] Definition. Extension Let \(\left(X, \leq_{X}\right)\) and \(\left(Y, \leq_{Y}\right)\) be linearly ordered sets. We say \(\left(Y, \leq_{Y}\right)\) is an extension of \(\left(X, \leq_{X}\right)\) if \(X \subseteq Y\) and, for all \(x_{1}, x_{2}\) in \(X\) , \[x_{1} \leq_{X} \quad x_{2} \quad \text { iff } \quad x_{1} \leq_{Y} x_{2} .\] THEOREM 8.21. Let \((X, \leq)\) be an extension of \((\mathbb{Q}, \leq)\) . If \((X, \leq)\) is order-complete and \(\mathbb{Q}\) is dense in \(X\) , then \(X\) is uncountable.
Proof. Suppose that \(X\) is a countable order-complete extension of \(\mathbb{Q}\) and that \(\mathbb{Q}\) is dense in \(X\) .
Let the sequence \(\left\langle a_{n} \mid n \in \mathbb{N}\right\rangle\) be a bijection from \(\mathbb{N}\) to \(X\) . Observe that the sequence imposes an ordering on \(X\) . Let \(\preceq\) be defined on \(X\) by \[(\forall m, n \in \mathbb{N}) a_{m} \preceq a_{n} \Longleftrightarrow m \leq n .\] That is, for any \(x, y \in X, x \preceq y\) if \(x\) appears in the sequence \(\left\langle a_{n}\right\rangle\) before \(y\) . Then \(\preceq\) is a well-ordering of \(X\) .
Given \(Y \subseteq X\) and \(y_{0} \in Y\) , we say that \(y_{0}\) is the \(\preceq\) -minimal element of \(Y\) if \[(\forall x \in Y) y_{0} \preceq x .\] So every subset of \(X\) has a \(\preceq\) -minimal element.
We shall define two subsequences of \(\left\langle a_{n}\right\rangle\) , called \(\left\langle a_{f(n)}\right\rangle\) and \(\left\langle a_{g(n)}\right\rangle\) , so that for any \(n \in \mathbb{N}\)
(1) \(f(n+1)>g(n)\)
(2) \(g(n)>f(n)\)
(3) \(a_{f(n+1)}\) is the \(\preceq\) -minimal element of the set \[\left\{y \in X \mid a_{f(n)}<y<a_{g(n)}\right\}\] (4) \(a_{g(n+1)}\) is the \(\preceq\) -minimal element of the set \[\left\{y \in X \mid a_{f(n+1)}<y<a_{g(n)}\right\} .\] We define the subsequences by recursion using the sequence \(\left\langle a_{n}\right\rangle\) to carefully control the construction. This argument is called a backand-forth argument. Given finite sequences of length \(N\) satisfying the properties enumerated above, we define \(a_{f(N+1)}\) subject to constraints imposed by \(a_{f(N)}\) and \(a_{g(N)}\) . We then define \(a_{g(N+1)}\) subject to constraints imposed by \(a_{f(N+1)}\) and \(a_{g(N)}\) . We then define \(a_{f(N+2)}, a_{g(N+2)}\) , and so on.
Let \(f(0)=0\) . So \(a_{f(0)}=a_{0}\) . Let \(g(0)\) be the smallest integer \(n\) such that \(a_{0}<a_{n}\) . Note that this is equivalent to defining \(g(0)\) so that \(a_{g(0)}\) is the \(\preceq\) -minimal element of \(X\) greater than \(a_{0}\) . Assume we have defined finite subsequences \(\left\langle a_{f(n)} \mid n \leq N\right\rangle,\left\langle a_{g(n)} \mid n \leq N\right\rangle\) satisfying the order properties listed above. We shall define \(a_{f(N+1)}\) and \(a_{g(N+1)}\) satisfying the ordering properties listed above. The set \(X\) contains the rational numbers and since \(\mathbb{Q}\) is dense in \(X\) , there is an element of \(X\) , \(x\) , such that \[a_{f(N+1)}<x<g_{(N+1)} .\] Let \(a_{f(N+1)}\) be the \(\preceq\) -minimal element of \(X\) such that \[a_{f(N)}<a_{f(N+1)}<a_{g(N)} .\] Since \(\preceq\) is a well-ordering of \(X, f(N+1)\) is well-defined. We let \(a_{g(N+1)}\) be the \(\preceq\) -minimal element of \(X\) such that \[a_{f(N+1)}<a_{g(N+1)}<a_{g(N)} .\] By our previous discussion, \(g(N+1)\) is well-defined. Observe that for any \(m, n \in \mathbb{N}\) , \[a_{f(m)}<a_{g(n)} .\] Therefore the increasing sequence \(\left\langle a_{f(n)} \mid n \in \mathbb{N}\right\rangle\) is bounded above, and by Lemma 8.5, the sequence converges to its least upper bound, \(a\) .
For any \(n \in \mathbb{N}\) , \[a_{f(n)}<a<a_{g(n) .} .\] So \(a\) is not a term of either subsequence. We show that \(a\) is not a term in the sequence \(\left\langle a_{n}\right\rangle\) . Suppose by way of contradiction that \(a=a_{n}\) for some \(n \in \mathbb{N}\) . Since \(f(0)=0, n \neq 0\) . Let \[Y=(f[\mathbb{N}] \cup g[\mathbb{N}]) \cap\ulcorner n\urcorner .\] Then \(Y \neq \emptyset\) is finite, and has a maximal element.
If the maximal element of \(Y\) is \(f(0)\) , then for every \(1 \leq k<n\) , we must have \(a_{k}<a_{0}\) . But then \(g(0)\) would be \(n\) , which contradicts the fact that \(n\) is not in the range of \(g\) .
If the maximal element of \(Y\) is \(f(m+1)\) for some \(m\) , then \(g(m+1)>\) \(n\) , and \[f(m+1)<n<g(m+1) .\] However \[a_{f(m+1)}<a_{n}<a_{g(m+1)}<a_{g(m) .} .\] This is impossible since \(a_{g(m+1)}\) is the \(\preceq\) -minimal element of \(X\) in the open interval \(\left(a_{f(m+1)}, a_{g(m)}\right)\) .
If the maximal element of \(Y\) is \(g(m)\) for some \(m\) , then \(f(m+1)>n\) and \[g(m)<n<f(m+1) .\] However \[a_{f(m)}<a_{f(m+1)}<a_{n}<a_{g(m)} .\] This is impossible since \(a_{f(m+1)}\) is the \(\preceq\) -minimal element of \(X\) in the open interval \(\left(a_{f(m)}, a_{g(m)}\right)\) . So \(a\) is not a term in the sequence \(\left\langle a_{n}\right\rangle\) . Therefore there is no bijection from \(\mathbb{N}\) to \(X\) , and \(X\) is uncountable.
By Exercise 8.20, \(\mathbb{Q}\) is dense in \(\mathbb{R}\) . As the set of real numbers is order-complete by the least upper bound theorem, we get:
COROLLARY 8.22. The set of real numbers is uncountable.
THEOREM 8.23. Let \(\left(X, \leq_{X}\right)\) be an order-complete extension of \(\mathbb{Q}\) in which \(\mathbb{Q}\) is dense, and such that \(X\) has no maximal or minimal element. Then there is an order-preserving bijection from \(\mathbb{R}\) onto \(X\) that is the identity on \(\mathbb{Q}\) .
Proof. Let us define a map \(f: \mathbb{R} \rightarrow X\) . If \(q \in \mathbb{Q}\) , define \(f(q)=q\) . If \(\alpha \in \mathbb{R} \backslash \mathbb{Q}\) , define \(f(\alpha)\) to be the least upper bound in \(X\) of \(\{q \in \mathbb{Q} \mid q \leq \alpha\}\) . The function \(f\) is well-defined, because \(X\) has the Least Upper Bound Property. It is injective, because if \(\alpha \neq \beta\) , there are rational numbers between \(\alpha\) and \(\beta\) .
To show \(f\) is onto, suppose \(x \in X\) . Define \(\alpha \in \mathbb{R}\) to be the least upper bound in \(\mathbb{R}\) of \(\left\{q \in \mathbb{Q} \mid q \leq_{X} x\right\}\) . Then \(f(\alpha)=x\) .
Finally, \(f\) is order-preserving because if \(\alpha \leq \beta\) , then \(f(\beta)\) is defined as the least upper bound of a superset of the set whose least upper bound is \(f(\alpha)\) , and so \(f(\alpha) \leq_{X} f(\beta)\) .
REMARK. What happens if we drop the requirement that \(X\) have no maximal or minimal element?