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 hence infinitely many others), and complete, i.e., which "lacks gaps" in the sense that every nonempty subset with an upper bound has a least upper bound. More symbolically:
Unlike the standard real line, a linear continuum may be bounded on either side: for example, any (real) closed interval is a linear continuum.
Examples in addition to the real numbers:
π1 (x, y) = x
This map is known as the projection map. The projection map is continuous (with respect to the product topology on I × I) and is surjective. Let A be a nonempty subset of I × I which is bounded above. Consider π1(A). Since A is bounded above, π1(A) must also be bounded above. Since, π1(A) is a subset of I, it must have a least upper bound (since I has the least upper bound property). Therefore, we may let b be the least upper bound of π1(A). If b belongs to π1(A), 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 (b × I) ∩ 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(A), 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(A) than b, contradicting the unique property of b.
A =
of the set of rational numbers. Even though this set is bounded above by any rational number greater than (for instance 3), it has no least upper bound in the rational numbers.[2] (Specifically, for any rational upper bound r >, r/2 + 1/r is a closer rational upper bound; details at .)
A = (−∞, 0) ∪ (0, +∞)
is not a linear continuum. Property b) is trivially satisfied. However, if B is the set of negative real numbers:
B = (−∞, 0)
then B is a subset of A which is bounded above (by any element of A greater than 0; for instance 1), but has no least upper bound in B. Notice that 0 is not a bound for B since 0 is not an element of A.
S = Z− ∪ A.
Then S satisfies neither property a) nor property b). The proof is similar to the previous examples.
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. (Munkres explains the second part of the proof in [3])
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:
A = (−∞, y)
B = (x, +∞)
These sets are disjoint (If a is in A, a < y so that if a is in B, a > x and a < y which is impossible by hypothesis), nonempty (x is in A and y is in B) and open (in the order topology), 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 (b, +∞) where b is an upper bound for C. Then D is open (since it is the union of open sets), and closed (if a is not in D, then a < b for all upper bounds b of C so that we may choose q > a such that q is in C (if no such q exists, a is the least upper bound of C), then an open interval containing a may be chosen that doesn't intersect D). Since D is nonempty (there is more than one upper bound of D for if there was exactly one upper bound s, s would be the least upper bound. Then if b1 and b2 are two upper bounds of D with b1 < b2, b2 will belong to D), D and its complement together form a separation on X. This contradicts the connectedness of X.