Linear Continuum - Topological Properties

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 (notice the 'if and only if' part). We will prove one implication, and leave the other one as an exercise. (Munkres explains the second part of the proof)

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 = (-∞, 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.