# Topology/Linear Continuum

An ordered set, *S*, is said to be a **linear continuum** if it satisfies the following properties:

a) *S* has the least upper bound property

b) 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. Linear continua are particularly important in the field of topology where they can be used to verify whether a ordered set given the order topology is connected or not.

## Examples edit

1. The ordered set of real numbers, **R**, with its usual order is a linear continuum. Property b) is trivial, and property a) is simply a reformulaton of the completeness axiom.

2. The set of rational numbers is not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset:

*A*= {*x*|*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.

3. The 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 that it has a maximum. 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.

4. The ordered set *A* of nonzero real numbers:

*A*= (-∞, 0) U (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 *A*. Notice that 0 is not a bound for *B* since 0 is not an element of *A*.

5. Let Z_{-} denote the set of negative integers and let A = (0,5) ∪ (5,+∞). Let:

- S = Z
_{-}U A

Then S satisfies neither property a) nor property b). The proof is similar to examples 3 and 4.

6. 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, π_{1} : I × *I* -> *I* by:

- π
_{1}(*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 π_{1}(*A*). Since *A* is bounded above, π_{1}(*A*) must also be bounded above. Since, π_{1}(*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 π_{1}(*A*). If *b* belongs to π_{1}(*A*), then *b* × *I* will intersect *A* at say *b* × *c* for some *c* isin; *I*. We merely choose the maximum c satisfying the property and *b* × *c* will be the desired least upper bound of *A * (note that: since *b* × *I* has the same [order type of I, the set (*b* × *I*) ∩ *A*, will indeed have a least upper bound). If *b* doesn't belong to π_{1}(*A*), then *b* × 0 is the least upper bound of *A*. for if *d* < *b*, and *d* × *e* is a least upper bound of *A*, then *d* would be a smaller upper bound of π_{1}(*A*) than *b*, contradicting the unique property of *b*.

## Topological properties of linear continua edit

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 iff it is a linear continua (notice the 'if and only if' part). 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 = (-∞, 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, +infinity) 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 an 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 b_{1} and b_{2} are two upper bounds of D with b_{1} < b_{2}, b_{2} will belong to D), D and its complement together form a separation on X. This contradicts the connectedness of X.

### Applications of the theorem edit

1. Notice that 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. Notice how 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 interesting example of a topological space that is a linear continuum, see long line (in Wikipedia).

## Exercises edit

1. a)* Prove that if X is a linear continuum, then X is connected.

b) Give an example of a set that satisfies property b) in the definition of a linear continuum, but isn't connected

2. a) Is a locally connected, ordered set given the order topology a linear continuum?

b) Does the converse of a) hold?

3. Show that if Y is a well ordered set, then Y X [0,1) is a linear continuum.

4*. a) Every path connected ordered set given the order topology is a linear continuum, but is every linear continuum necessarily path connected? (Hint: Why not check the dictionary order topology?)

b) The reader is referred to exercise 4 of the next section for the answer to this problem. However, the reader is encouraged not to see exercise 4 in the next section unless he/she has solved the problem. Note that * is used in 4.a) to indicate difficulty. You must prove your claim in 4.a) !

5.a) Determine whether or not an ordered set with the order topology satisfying b) has finitely many components.

b) Notice that property a) of the linear important is crucially important in order to determine the connectedness of an ordered set given the order topology. Show that if X is an ordered set with the order topology that only satisfies property b) in the definition of a linear continuum, then X may be totally disconnected. Give an example of such an ordered set X.

6. Is the following result true:

An ordered set X has the order topology equal to the discrete topology on X, if and only if X doesn't satisfy property b) in the definition of a linear continuum.

If no, does at least one implication hold and which one is it?