The ordered set of real numbers, R, with its usual order is a linear continuum, and is the archetypal example. Property b) is trivial, and property a) is simply a reformulaton of the completeness axiom.
The set I × I is trivial. To check property a), we define a map, π1 : I × I → I by:
This map is known as the projection map. The projection map is continuous and is surjective. Let A be a nonempty subset of I × I which is bounded above. Consider π1. Since A is bounded above, π1 must also be bounded above. Since, π1 is a subset of I, it must have a least upper bound. Therefore, we may let b be the least upper bound of π1. If b belongs to π1, then b × I will intersect A at say b × c for some c ∈ I. Notice that since b × I has the same order type of I, the set ∩ A will indeed have a least upper bound b × c', which is the desired least upper bound for A. If b does not belong to π1, then b × 0 is the least upper bound of A, for if d < b, and d × e is an upper bound of A, then d would be a smaller upper bound of π1 than b, contradicting the unique property ofb.
Non-examples
The set of rational numbers is not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset:
The set of non-negative integers with its usual order is not a linear continuum. Property a) is satisfied is not. Indeed, 5 is a non-negative integer and so is 6, but there exists no non-negative integer that lies strictly between them.
The ordered set A of nonzero real numbers:
Let Z− denote the set of negative integers and let A = ∪. Let:
Topological properties
Even though linear continua are important in the study of ordered sets, they do have applications in the mathematical field of topology. In fact, we will prove that an ordered set in the order topology is connected if and only if it is a linear continuum. We will prove one implication, and leave the other one as an exercise. Theorem Let X be an ordered set in the order topology. If X is connected, then X is a linear continuum. Proof: Suppose, x is in X and y is in X where x < y. If there exists no z in X such that x < z < y, consider the sets: A = B = These sets are disjoint, nonempty and open and their union is X. This contradicts the connectedness of X. Now we prove the least upper bound property. If C is a subset of X that is bounded above and has no least upper bound, let D be the union of all open rays of the form where b is an upper bound for C. Then D is open, and closed. Since D is nonempty, D and its complement together form a separation on X. This contradicts the connectedness of X.
Applications of the theorem
1. Since the ordered set: A = U is not a linear continuum, it is disconnected. 2. By applying the theorem just proved, the fact that R is connected follows. In fact any interval in R is also connected. 3. The set of integers is not a linear continuum and therefore cannot be connected. 4. In fact, if an ordered set in the order topology is a linear continuum, it must be connected. Since any interval in this set is also a linear continuum, it follows that this space is locally connected since it has a basis consisting entirely of connected sets. 5. For an example of a topological space that is a linear continuum, see long line.
