Linear continuum
In the mathematical field of order theory, a continuum or linear continuum is a generalization of the real line.
Formally, a linear continuum is a linearly ordered set S of more than one element that is densely ordered, i.e., between any two distinct elements there is another, and complete, i.e., which "lacks gaps" in the sense that every nonempty subset with an upper bound has a least upper bound in the set. More symbolically:
- S has the least upper bound property, and
- For each x in S and each y in S with x < y, there exists z in S such that x < z < y
Unlike the standard real line, a linear continuum may be bounded on either side: for example, any closed interval is a linear continuum.
Examples
- 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 reformulation of the completeness axiom.
- sets which are order-isomorphic to the set of real numbers, for example a real open interval, and the same with half-open gaps
- the affinely extended real number system and order-isomorphic sets, for example the unit interval
- the set of real numbers with only +∞ or only −∞ added, and order-isomorphic sets, for example a half-open interval
- the long line
- The set I × I is trivial. To check property a), we define a map, π1 : I × I → I by
Non-examples
- The ordered set Q of rational numbers is not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset
- The ordered 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
Theorem
Let X be an ordered set in the order topology. If X is connected, then X is a linear continuum.
Proof:
Suppose that x and y are elements of X with x < y. If there exists no z in X such that x < z < y, consider the sets:
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
- Since the ordered set A = U is not a linear continuum, it is disconnected.
- By applying the theorem just proved, the fact that R is connected follows. In fact any interval in R is also connected.
- The set of integers is not a linear continuum and therefore cannot be connected.
- 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.
- For an example of a topological space that is a linear continuum, see long line.