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Math 403 Lecture Notes for Week 1
1 Ordered Fields
Definition 1 (Field Axioms). A field is a set X together with two binary functions + and · defined on X which satisfy the following properties:
- x + y = y + x
- (x + y) + z = x + (y + z)
- There is an element, denoted 0, such that x + 0 = x for all x.
- For all x there is a w, denoted −x, such that x + w = 0.
- x · y = y · x
- (x · y) · z = x · (y · z)
- There is an element, denoted 1, such that x · 1 = x for all x.
- For all x 6 = 0 there is a w, denoted x−^1 , such that x · w = 1.
- x · (y + z) = x · y + x · z
The real numbers R with the usual addition and multiplication is a field. So are the rationals Q and the complex numbers C with the usual operations.
Example 2. We can define a somewhat more exotic field F 2 as follows. F 2 = { 0 , 1 } has two elements, with + and · defined as follows:
Definition 3. A strictly totally ordered set consists of a set X and a binary relation < which satisfies:
- x ≮ x
- For all x and y, exactly one of the three possibilities holds: x < y, x = y, y < x.
- If x < y and y < z then x < z.
Definition 4 (Order Axioms). An ordered field consists of a field (X, +, ·) and a set P ⊆ X of positive elements satisfying the following:
- If x ∈ P and y ∈ P , then x + y ∈ P.
- If x ∈ P and y ∈ P , then x · y ∈ P.
- If x ∈ P then −x /∈ P.
- For each x, exactly one of the three possibilities holds: x = 0, x ∈ P , −x ∈ P.
Example 5. 1. The reals with the set P = {x : x > 0 } is an ordered field.
- The reals with the set P = {x : x < 0 } is not an ordered field since − 1 ∈ P but (−1) · (−1) ∈/ P.
- The rationals Q with P = {x ∈ Q : x > 0 } is an ordered field.
- The field F 2 cannot be ordered, i.e. there is no set P which makes it an ordered field. To see this, note that −1 = 1, so if 1 ∈ P then we need − 1 ∈/ P , i.e. 1 ∈/ P , whereas if 1 ∈/ P we need − 1 ∈ P , i.e. 1 ∈ P.
Definition 6. We can define a strict totally ordering on an ordered field by setting x < y if and only if y − x ∈ P
We write x ≤ y for (x = y or x < y).
Note 7. We can restate the ordering axioms in terms of the order < as follows:
- If x < y and z < w then x + z < y + w.
- If 0 < x < y and 0 < z < w then x · z < y · w.
- Exactly one of x < y, x = y, y < x.
2 Bounded Sets
Definition 8. Let A be a subset of an ordered field.
- We say that b is an upper bound of A if x ≤ b for all x ∈ A.
- We say that c is a lower bound of A if c ≤ x for all x ∈ A.
- We say that b is a least upper bound of A if b is an upper bound of A and for any other upper bound c of A we have b ≤ c.
- We say that c is a greatest lower bound of A if c is a lower bound of A and for any other lower bound d of A we have d ≤ c.
- We say that b is the greatest element of A if b is an upper bound of A and b ∈ A.
- We say that c is the least element of A if c is a lower bound of A and c ∈ A.
Then S is non-empty since it includes 0 and has an upper bound, namely x. Hence S has a least upper bound y. The element y − 12 is then not an upper bound for S, so there is some integer k ∈ S with y − 12 < k. But then y < k + 12 < k + 1, so k + 1 is not in S. Since k + 1 is an integer, this means that x < k + 1, so setting n = k + 1 works. 2
Corollary 17. The rationals are dense in a complete ordered field, i.e. if x < y then there is a rational r with x < r < y.
Proof: First suppose x ≥ 0. By the Axiom of Archimedes, there is an integer q with q > (y − x)−^1 , so y − x > (^1) q. Consider the set S with
S = {n : n is an integer with y ≤
n q
S is non-empty and bounded below by y · q, so it has a greatest lower bound p. Since S consists only of integers, p is in fact an element of S, so p is the least element of S. This means that
p − 1 q
< y ≤
p q
We also have x = y − (y − x) < pq − (^1) q = p− q 1. So letting r = p− q 1 we have x < r < y. If x < 0 we can find an integer n with n > −x, so n + x > 0. As above, we then find a rational r with n + x < r < n + y. The number r − n is then a rational with x < r − n < y. 2
3 Infinite Decimals
For definiteness, we will define the reals to be infinite decimals. By an infinite decimal, we mean a number of the form
n.d 0 d 1 d 2 · · ·
or the negation of such a number, where n and each di is an integer and 0 ≤ di ≤ 9 for each i ≥ 1. This sequence represents the real number
n +
i≥ 1
di 10 i
These are added and multiplied with the expected carrying of digits. The one point to be careful about is that we must identify eventually 9 sequences and eventually 0 sequences, such as 0. 999 · · · = 1. 000 · · ·. With this understanding, the collection of infinite decimals forms a complete ordered field.
4 Limits
Definition 18. A sequence is a list of real numbers indexed by the positive integers. We use the notation 〈xn〉n≥ 1 to denote the sequence x 1 , x 2 , x 3 ,.. ..
Definition 19 (N - definition of limit). Let 〈xn〉n≥ 1 be a sequence and y a real number. We say that the sequence has limit y, and write limn→∞ xn = y if: For any > 0 there is a natural number N such that |xn − y| < for all n ≥ N.
Definition 20. Let 〈xn〉n≥ 1 be a sequence. We say that the sequence has limit ∞, and write limn→∞ xn = ∞ if: For any k there is a natural number N such that xn > k for all n ≥ N. We define limit −∞ similarly.
Note 21. We shall try to be careful about what we mean when we say a sequence has a limit. We shall generally say a sequence has a limit as a real number to indicate that there is a real number y which is the limit of the sequence. We will say a sequence has a limit as an extended real number if we also allow the limit to be ±∞.
Example 22. Consider the sequence with x + n = (^1) n. Then limn→∞ = 0. To see this, let > 0 be given. We need to find an N such that |xn − 0 | < for all n ≥ N , i.e. (^) n^1 < . This is equivalent to ensuring n > (^1) , so choosing N > (^1) suffices.
Definition 23. We say that a sequence 〈xn〉n≥ 1 is bounded above if the set {xn : n ≥ 1 } is bounded above. We define bounded below similarly.
Definition 24. We say a sequence is monotone increasing if xn ≤ xn+1 for all n ≥ 1. We define monotone decreasing similarly.
Proposition 25. Let 〈xn〉n≥ 1 be a sequence which is bounded above and mono- tone increasing. Then 〈xn〉n≥ 1 has a limit as a real number.
Proof: Consider the set {xn : n ≥ 1 }. It is non-empty and bounded above, so it has a least upper bound y. We claim that limn→∞ xn = y. Let > 0 be given. We want to find an N as required in the definition of limit. Suppose there is none. Then for any N there is an n ≥ N with |xn − y| ≥ . This means xn ≤ y − . Since the sequence is monotone increasing, we have xN ≤ xn ≤ y − as well. Since this holds for any N , we have that y − is an upper bound for the sequence, contradicting that y was a least upper bound. 2
Note 26. The same is true for a sequence which is bounded below and monotone decreasing.
Note 27. Both conditions in the proposition are necessary. For instance, the sequence 1, − 1 , 1 , − 1 , · · · is bounded, but has no limit. The sequence with xn = n is monotone but not bounded above, and has no limit as a real number. Note, though, that a monotone increasing sequence always has a limit as an extended real number, since if it is not bounded above it has limit ∞.
