Linear continuum
Encyclopedia
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 members there is another, and which "lacks gaps" in the sense that every non-empty subset with an upper bound has a least upper bound. More symbolically:
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 an ordered set
given the order topology
is connected
or not.
Examples in addition to the real numbers:
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 doesn't 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.
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.
is not a linear continuum. Property b) is trivially satisfied. However, if B is the set of negative real numbers:
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.
Then S satisfies neither property a) nor property b). The proof is similar to the previous examples.
, 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, +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 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.
:
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
.
Order theory
Order theory is a branch of mathematics which investigates our intuitive notion of order using binary relations. It provides a formal framework for describing statements such as "this is less than that" or "this precedes that". This article introduces the field and gives some basic definitions...
, a continuum or linear continuum is a generalization of the real line
Real line
In mathematics, the real line, or real number line is the line whose points are the real numbers. That is, the real line is the set of all real numbers, viewed as a geometric space, namely the Euclidean space of dimension one...
.
Formally, a linear continuum is a linearly ordered set S of more than one element that is densely ordered
Dense order
In mathematics, a partial order ≤ on a set X is said to be dense if, for all x and y in X for which x In mathematics, a partial order ≤ on a set X is said to be dense if, for all x and y in X for which x...
, i.e., between any two members there is another, and which "lacks gaps" in the sense that every non-empty subset with an upper bound has a least upper bound. More symbolically:
a) S has the least-upper-bound property
Least-upper-bound property
In mathematics, the least-upper-bound property is a fundamental property of the real numbers and certain other ordered sets. The property states that any non-empty set of real numbers that has an upper bound necessarily has a least upper bound ....
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
Subset
In mathematics, especially in set theory, a set A is a subset of a set B if A is "contained" inside B. A and B may coincide. The relationship of one set being a subset of another is called inclusion or sometimes containment...
of the set that is bounded above has a least upper bound. Linear continua are particularly important in the field of topology
Topology
Topology is a major area of mathematics concerned with properties that are preserved under continuous deformations of objects, such as deformations that involve stretching, but no tearing or gluing...
where they can be used to verify whether an ordered set
Total order
In set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
given the order topology
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
is connected
Connected space
In topology and related branches of mathematics, a connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. Connectedness is one of the principal topological properties that is used to distinguish topological spaces...
or not.
Examples
- The ordered set of real numberReal numberIn mathematics, a real number is a value that represents a quantity along a continuum, such as -5 , 4/3 , 8.6 , √2 and π...
s, R, with its usual orderTotal orderIn set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
is a linear continuum, and is the archetypal example. Property b) is trivial, and property a) is simply a reformulaton of the completeness axiomCompleteness axiomIn mathematics the completeness axiom, also called Dedekind completeness of the real numbers, is a fundamental property of the set R of real numbers...
.
Examples in addition to the real numbers:
- sets which are order-isomorphicOrder isomorphismIn the mathematical field of order theory an order isomorphism is a special kind of monotone function that constitutes a suitable notion of isomorphism for partially ordered sets . Whenever two posets are order isomorphic, they can be considered to be "essentially the same" in the sense that one of...
to the set of real numbers, for example a real 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 intervalUnit intervalIn mathematics, the unit interval is the closed interval , that is, the set of all real numbers that are greater than or equal to 0 and less than or equal to 1...
- the set of real numbers with only +∞ or only -∞ added, and order-isomorphic sets, for example a half-open interval
- the long lineLong line (topology)In topology, the long line is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties. Therefore it serves as one of the basic counterexamples of topology...
- The set I × I (where × denotes the Cartesian productCartesian productIn mathematics, a Cartesian product is a construction to build a new set out of a number of given sets. Each member of the Cartesian product corresponds to the selection of one element each in every one of those sets...
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 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
Upper bound
In mathematics, especially in order theory, an upper bound of a subset S of some partially ordered set is an element of P which is greater than or equal to every element of S. The term lower bound is defined dually as an element of P which is lesser than or equal to every element of S...
(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
Order type
In mathematics, especially in set theory, two ordered sets X,Y are said to have the same order type just when they are order isomorphic, that is, when there exists a bijection f: X → Y such that both f and its inverse are monotone...
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 doesn't 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.
Non-examples
- The set of rational numberRational numberIn mathematics, a rational number is any number that can be expressed as the quotient or fraction a/b of two integers, with the denominator b not equal to zero. Since b may be equal to 1, every integer is a rational number...
s is not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset:
- A = { x | x < √ }
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.
- The set of non-negative integerIntegerThe integers are formed by the natural numbers together with the negatives of the non-zero natural numbers .They are known as Positive and Negative Integers respectively...
s 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.
- The ordered setTotal orderIn set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
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 A. 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 setsTotal order
In set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
, they do have applications in the mathematical field of topology
Topology
Topology is a major area of mathematics concerned with properties that are preserved under continuous deformations of objects, such as deformations that involve stretching, but no tearing or gluing...
. In fact, we will prove that an ordered set
Total order
In set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
in the order topology
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
is connected
Connected space
In topology and related branches of mathematics, a connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. Connectedness is one of the principal topological properties that is used to distinguish topological spaces...
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
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
. If X is connected
Connected space
In topology and related branches of mathematics, a connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. Connectedness is one of the principal topological properties that is used to distinguish topological spaces...
, 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
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
) 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
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
of the form (b, +infinity) where b is an upper bound for C. Then D is open
Open set
The concept of an open set is fundamental to many areas of mathematics, especially point-set topology and metric topology. Intuitively speaking, a set U is open if any point x in U can be "moved" a small amount in any direction and still be in the set U...
(since it is the union of open sets), and closed
Closed set
In geometry, topology, and related branches of mathematics, a closed set is a set whose complement is an open set. In a topological space, a closed set can be defined as a set which contains all its limit points...
(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
Order topology
In mathematics, an order topology is a certain topology that can be defined on any totally ordered set. It is a natural generalization of the topology of the real numbers to arbitrary totally ordered sets...
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
Separated sets
In topology and related branches of mathematics, separated sets are pairs of subsets of a given topological space that are related to each other in a certain way....
on X. This contradicts the connectedness
Connectedness
In mathematics, connectedness is used to refer to various properties meaning, in some sense, "all one piece". When a mathematical object has such a property, we say it is connected; otherwise it is disconnected...
of X.
Applications of the theorem
1. Notice that since the ordered setTotal order
In set theory, a total order, linear order, simple order, or ordering is a binary relation on some set X. The relation is transitive, antisymmetric, and total...
:
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
Connected space
In topology and related branches of mathematics, a connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. Connectedness is one of the principal topological properties that is used to distinguish topological spaces...
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
Locally connected space
In topology and other branches of mathematics, a topological space X islocally connected if every point admits a neighbourhood basis consisting entirely of open, connected sets.-Background:...
since it has a basis consisting entirely of connected sets.
5. For an interesting example of a topological space
Topological space
Topological spaces are mathematical structures that allow the formal definition of concepts such as convergence, connectedness, and continuity. They appear in virtually every branch of modern mathematics and are a central unifying notion...
that is a linear continuum, see long line
Long line (topology)
In topology, the long line is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties. Therefore it serves as one of the basic counterexamples of topology...
.