Linear continuum

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In the

Formally, a linear continuum is a

1. S has the
   least upper bound property
   , and
2. For each x in S and each y in S with x < y, there exists z in S such that x < z < y

A set has the least upper bound property, if every nonempty subset of the set that is bounded above has a least upper bound in the set. Linear continua are particularly important in the field of topology where they can be used to verify whether an ordered set given the order topology is connected or not.^[1]

Unlike the standard real line, a linear continuum may be bounded on either side: for example, any (real)

Examples

• The ordered set of
  completeness axiom
  .

Examples in addition to the real numbers:

• sets which are
  open interval
  , and the same with half-open gaps (note that these are not gaps in the above-mentioned sense)
• 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 (where × denotes the Cartesian product and I = [0, 1]) in the lexicographic order is a linear continuum. Property b) is trivial. To check property a), we define a map, π₁ : I × I → I by

π₁ (x, y) = x

This map is known as the

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.

• 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

A = {x ∈ Q | x < √2}

of the set of rational numbers. Even though this set is bounded above by any rational number greater than √2 (for instance 3), it has no

least upper bound in the rational numbers.^[2] (Specifically, for any rational upper bound r > √2, r/2 + 1/r is a closer rational upper bound; details at Methods of computing square roots § Heron's method

.)

• The ordered set of non-negative integers with its usual order is not a linear continuum. Property a) is satisfied (let A be a subset of the set of non-negative integers that is bounded above. Then A is finite so it has a maximum, and this maximum is the desired least upper bound of A). On the other hand, property b) 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

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.

• Let Z₋ denote the set of negative integers and let A = (0, 5) ∪ (5, +∞). Let

S = Z₋ ∪ A.

Then S satisfies neither property a) nor property b). The proof is similar to the previous examples.

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. (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

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.

Applications of the theorem

1. Since the ordered set A = (−∞, 0) U (0,+∞) 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 (or ray) 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
   basis
   consisting entirely of connected sets.
5. For an example of a topological space that is a linear continuum, see long line.

See also

• Cantor-Dedekind axiom
• Order topology
• Least upper bound property
• Total order

References