...ield. S is some subset of all ordered pairs of elements of Y. , and (a, b) is an element of S with the notation p b and c > d, then a + c > b + d
• If a > b, and c > 0, then a * c > a * b
• If a > 0, then a -1 > 0
• If 0 < a and 0 < b, then 0 < a * b
• If a > b and b > c, then a > c
• If a ≥ b and b ≥ a, then a = b
• If a ≠ 0, then a 2 > 0
• 1 > 0
• If n is a natural number, then n > 0
I will prove a few of the properties that I think are the most important.
• For all a and b in Y one of the following statement must be true:
a < b, a = b, b < a
First of all if a, b belongs to Y, then their sum a + b belongs to Y. Also if a belongs to Y then one of the following relations holds: a is an element of Y, a = 0, -a is an element of Y. Therefore only one of the following holds: a –b belongs to Y, a – b = 0 or b – a = - (a – b ) belongs to Y.
• If a > b and c > d, then a + c > b + d
If a – b and c – d belong to Y, then (a + c) – (b + d) = (a – b) + (c – d) also belongs to Y.
• If a ≠ 0, then a2 > 0
If a, b belongs to Y, then so does their product a * b. Either a or – a belongs to Y. If a is an element of Y, then a2 =a * a or -a *-a also belongs to Y.
Dedekind Complete ordered field
As we have learnt that an upper bound is for S a subset of R and u is an upper bound of S is s < u for all s are elements of S. The least upper bound l, is an upper bound for S, if l is smaller or equal to any other upper bounds of S.
A Dedekind ordered field Y, is complete, if the following holds:
For all nonempty subset S of Y, S has more than one upper bound, then S has a least upper bound.
A complete order field is Archimedean. A field is Archimedean if for each x in F there exists a natural number n such that x < n. Therefore the following conditions hold:
• If a > 0 and b > 0, there is a natural number n such that n * a > b
• If b > 0, there is a natural number n such that 0 < 1/n < b
• If a > 0, there is a natural number n such that n – 1 ≤ a < n
Dedekind cuts
The number line is filled with rationals as closed as we want, between any two points of the number line there is a rational. We can define a real number by dividing the rationals into two sets A and B, A with whose to the left and B with those to the right. Therefore a real number is a Dedekind cut of the rationals.
A Dedekind cut of real rationals is a pair of sets ( A, B ) where its elements are all rationals and the following conditions hold:
• A and B are both nonempty sets.
• All rational numbers belongs to either A or B.
• No rational number belongs to both A and B.
• If a belongs to A and b belongs to B then a < b.
• A does not have a greatest member.
Defining real number with Dedekind cuts.
First of all divide the number line to define a real number by
A = {a an element of rationals: a ≤ 0} U {a an element of rationals: a > 0 and a 2 < 2}
B = {a an element of the rationals: a > 0 and a 2 > 2}.
This cut represents the real number √2. Sets A and B are both nonempty sets, suppose that p is an element of B. We cannot find p in A, therefore p belongs to B, and B is nonempty, and hence A is nonempty.
All rational numbers belongs to either A or B, A U B is the set of real numbers, and since B is the set of all rationals not belonging to A, hence they are mutually exclusive, therefore no rational number belongs to both A and B.
Since A and B are mutually exclusive element from B will not be in A, therefore an element from A will always be smaller than an element from B, since B is on the right of A on the number line. A does not have a greatest member, as you can always find another rational number between any two points on the number line.
Also since B has no least element, the point we are defining does not represent a rational.
If p is an element of A α, then p is less than α. If q is an element of A β q is less than β, so p + q is less than α + β. Hence the α + β Dedekind cuts A set can be define as A α + β ={p + q: p element of A α, q element of A β }, now by definition a Dedekind cut of β is a pair(A, B) of nonempty subsets of R, with A U B = R, such that α < β for all α an element of A and β an element of B. Hence we can say B α + β = QA α + β.
Now let us analyze the bullet points. According to the first point, we need to have both the sets A and B to be non empty. Suppose that p is an element of B α, and q of B β. Then if p + q = r + s, either r ≥ p or s ≥ q. So we cannot find an r in A α and an s in A β with p + q = r + s. Therefore p + q belongs to B and so B is nonempty. Hence A is nonempty as well.
It is clear from the definition that all the rational belongs to either A or B and that no rational belongs to both sets.
Fourth point suppose that d is an element of A and that e < d. then d = a + b for some a and b, with a an element of A α and b an element of A β. e = d + e – d = (a + e – d) + b. But a + e – d < a, so that (a + e – d) is an element of A α. Thus e an element of A. Hence if d is an element of A and f an element of B then d < f.
Lastly, suppose A has a greatest member c. Then c = a + b, where a is the greatest member of A α, b is the greatest member of A β. This contradicts the definition of A α and A β. Therefore A does not have a greatest member.
If p is an element of A α, then p is less than α. If q is an element of A β q is less than β, so p*q < α*β. Hence the α*β Dedekind cuts set A can be define as
Aα*β = {p*q: p element of A α, q element of A β} and set B define as Bα*β = {the set of rationals not belonging to Aα*β}.
Now looking at the bullet points again. Suppose that p is an element of B α, and q of B β. Then if p*q = r*s, either r ≥ p or s ≥ q. So we cannot find an r in A α and an s in A β with p*q = r*s. Therefore p*q belongs to A and so A is nonempty. It is evident than B is nonempty.
It is clear from the definition that all the rational belongs to either A or B and that no rational belongs to both sets.
Since we defined p and q are less than α and β, then p*q is less than α*β, therefore the fourth point is satisfied.
The last point, suppose A has a greatest member c. Then c = a*b, where a is the greatest member of A α and b is the greatest member of A β. This contradicts the definition of A α and A β. Therefore A does not have a greatest member therefore not bounded. Therefore (A, B) is a Dedekind cut.
Now I am going to define the zero element, θ. θ is define by the two cuts
A θ = {p element of Q: p < 0} and B θ = {q element of Q: q ≥ 0}.
Least upper bound
We can prove using the Dedekind cut that if Y is a nonempty set of real numbers which is bounded above, then from all the upper bounds we may pick a least upper bound.
Let a typical member of the set Y be Xi = (Ai, Bi) so that i changes we have all members of Y. Y is bounded above, so let (A,B) be an upper bound for Y.
Define A* and B* by A* = Ui Ai and B* = ∩i Bi.
We shall now show that (A*, B*) is a Dedekind cut of the rationals. Since any one of the Ai’s is non-empty, therefore A* is nonempty. Since (A, B) is an upper bound for Y, (Ai, Bi) ≤ (A, B) for all I; so that, for all i, A belongs to Ai. Thus A belongs to A*, so A* is non empty.
If a rational does not belong to A* then it does not belong to any of the Ai’s. So it belongs to all of the Bi’s and hence to B*. If a rational belongs to A* then it belongs to at least one of the Ai’s. Therefore it does not belong to at least one Bi. Hence it does not belong to B*.
If a belongs to A* and b belongs to B*, then a belongs to at least one Ai, say A1. b belongs to all the Bi’s, so in particular it belongs to B1. Thus a < b.
If A* has a greatest member then this member belongs to one of the Ai’s, say A1. It must be the greatest member of A1. This contradicts the fact that (A1, B1) is a Dedekind cut.
Thus (A*, B*) is a Dedekind cut and represents a real number. Since Ai belongs to A* for all i, (Ai, Bi) < (A*, B*) for all i, so that (A*, B*) is a...