Download Ordered Set: An order on a set S is a relation, denoted by < with the

Ordered Set: An order on a set S is a relation, denoted by < with the following
properties:
(i) If x, y ∈ S then one and only one of the following statements is true: x < y, x =
y y < x.
(ii) If x, y, z ∈ S such that x < y and y < z then x < z.
An Ordered Set is a set S in which an order is defined.
Definition: Suppose S is an ordered set, and E ⊂ S. If there exists a β ∈ S such
that x ≤ β for every x ∈ E we say that E is bounded above, and call β an upper
bound of E.
Lower bounds are defined similarly.
Definition: Suppose S is an ordered set, E ⊂ S and E is bounded above. Suppose
there exists an α ∈ S such that:
(i) α is an upper bound of E
(ii) If γ < α then γ is not an upper bound of E. Then, α is called the least upper
bound of E and write α = supE.
The greatest lower bound or infimum, of a set E that is bounded below, is an
element α ∈ S such that
(i) α is an lower bound of E
(ii) If γ > α then γ is not a lower bound of E. Then, α is called the greatest lower
bound of E and write α = inf E.
An ordered set S is said to have the least upper bound property if the following
is true:
If E ⊂ S, E is not empty, and E is bounded above, then supE exists in S.
Theorem 1: Suppose S is an ordered set with the least upper bound property,
B ⊂ S, B is not empty, and B is bounded below. Let L be the set of all lower bounds
of B. Then, α = supL exists in S and α = inf B.
Proof: Let L = { y ∈ S|y is a lower bound of B}.
Since B is bounded below, there exists at least one y ∈ S such that y is a lower bound
of B. Thus, L is a non empty subset of S.
And, since each y ∈ L is a lower bound of B, we have that, y ≤ x for every x ∈ B.
Thus, each x in B is an upper bound of L. Hence, L is bounded above.
Since S has the l.u.b. property, this implies sup L exists in S. Let α = supL.
Now, to show that α = infB.
We first show that α is a lower bound of B.
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That is to show that α ≤ x for every x ∈ B.
If not, there must be a γ ∈ B such that γ < α.
Since α is the least upper bound, γ cannot be an upper bound of L. Hence γ ∈
/ B.
(Why?)
This is a contradiction. Hence, α ≤ x for every x ∈ B and is hence α a lower bound
of B. (Thus, α ∈ L.)
Now, to show that α is the greatest lower bound of B.
If there is a β ∈ S such that α < β, then such a β ∈
/ L. (Why?) Thus, α is an lower
bound of B but no β > α is a lower bound of B. Hence, α = inf B.
Fields: A field is a set F with two operations, called addition and multiplication that
satisfy the following axioms:
Axioms for Addition:
(A1) If x, y ∈ F then their sum x + y ∈ F .
(A2) Addition is commutative: x + y = x + y for all x, y ∈ F .
(A3) Addition is associative: (x + y) + z = x + (y + z), for all x, y, z ∈ F.
(A4) F contains an element 0 such that 0 + x = x for every x ∈ F .
(A5) To every x ∈ F corresponds an element −x ∈ F such that x + (−x) = 0.
Axioms for multiplication:
(M1) If x ∈ F and y ∈ F, then their product xy is in F .
(M2) Multiplication is commutative: xy = yx for all x, y ∈ F .
(M3) Multiplication is Associative: (xy)z = x(yz) for all x, y, z ∈ F .
(M4) F contains an element 1 6= 0 such that 1x = x for every x ∈ F .
(M5) If x ∈ F and x 6= 0 then there exists an element x1 ∈ F such that x.( x1 ) = 1.
Distributive Law: x(y + z) = xy + xz, for all x, y, z ∈ F .
Proposition 1: The axioms for addition imply the following statements:
(a) If x + y = x + z then y = z.
(b) If x + y = x then y = 0.
(c) If x + y = 0 then y = −x.
(d) −(−x) = x.
Proposition 2: The axioms for multiplication imply the following statements:
(a) If x 6= 0 and xy = xz then y = z.
(b) If x 6= 0 and xy = x then y = 1.
(c) If x 6= 0 and xy = 1 then y = x1 .
(d) −(−x) = x.
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Proposition 3: The field axioms imply the following statements for x, y, z ∈ F .
(a) 0x = 0.
(b) If x 6= 0 and y 6= 0 then xy 6= 0.
(c) (−x)y = −(xy) = x(−y).
(d) (−x)(−y) = xy.
Definition: An ordered field is a field F is an ordered set such that:
(i) x + y < x + z if x, y, z ∈ F and y < z
(ii) xy > 0 if x ∈ F, y ∈ F, x > 0 and y > 0. If x > 0 then x is called positive and if
x < 0 it is called negative.
Proposition 4: The following statements are true in any ordered field.
(a) If x > 0 then −x < 0 and vice versa.
(b) If x > 0 and y < z the xy < xz.
(c) If x < 0 and y < z then xy > xz.
(d) If x 6= 0 then x2 > 0. In particular, 1 > 0.
(e) If 0 < x < y then 0 < y1 < x1 .
Theorem 1: There exists an ordered field IR which has the least upper bound property. IR contains Q as a subfield.
Theorem 2:
(a) (Archimedean Property) If x, y ∈ IR and x > 0 then ethere exists a positive
integer n such that nx > y.
(b) (Density of Rationals in IR): If x, y ∈ IR and x < y then there exists a p ∈ Q
such that x < p < y.
(c) For every real x > 0 and evevry integer n > 0 there is one and only one positive
rel y such that y n = x.
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(d) If a, b are positive real numbers and n is a positive integer, then (ab) n = a n b n .
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