MTH104 Handouts Old

1 August 29, 2013

Hard: #22 in Chapter 1. Consider a pile of sand principle. You wish to take away part of it to make it smaller than some number ϵ. Suppose you can take away 5% of what remains on any given day. Eventually you will reduce the pile of sand to less than ϵ. Today we will cover section 1 and part of section 2.

1.1 Notation

∀ is read “for each”, do NOT use “for all”. ∃ is read “there exists”. ⇒ is read “implies”. Then A ⇒ B. Note that the principle of explosion exists. Set notation: we use capital letters to denote sets, and small letters to denote the elements of the set (e.g. x ∈ X ). Then x is called an element (or a point of such a set). Then {x ∈ X : . . . }

is read “the set of all x in X such that. . . ”. Common sets:

• ∅ is the empty set.
• N = {1, 2, 3, . . . } is the set of natural numbers.
• Z is the set of integers.
• Q is the set of rational numbers. We can write this as

Q =  p q : p ∈ N ∪{0} , q ∈ Z \ {0}

This is a terrible definition, probably a better one is

Q =  p q : p ∈ Z, q ∈ N

Note that N ⊂ Z ⊂ Q. Note that in this course, ⊂ will mean what is usually meant by

We will assume all the nice properties of these sets. The subject of the course, however is R.

1.2 The Real Numbers

In most books, we accept properties of R as axioms. We will instead construct R and prove that the “axioms” are true. The most important property of the real numbers is the least upper bound property.

Fact 1.1 (Least Upper Bound). If S is a nonempty set of real numbers and there exists an upper bound for S, then there exists a least upper bound in R for S.

Note: we say a upper bound for S, not of S; the latter suggests the upper bound is contained in S.

Definition 1.2. An upper bound for a set S is a real M such that M ≥ s∀s ∈ S. Example 1.3. Let S = {r ∈ Q : r < 2}. Then an upper bound for S is 9000, but the least upper bound is 2.

Note that the least upper bound property is NOT true for the rationals. For example, consider

This has no least upper bound in the rationals, but it is 2 in the reals.

1.3 Constructing the Reals

Theorem 1.4 (Irrationality of the Square Root of 2). For every rational p q ∈ Q, we

have p q 2 = 2.

Proof. Suppose for contradiction that p q 2 = 2 where (p, q) = 1. In particular, p and q are both not even. For the first case, assume p is odd. But now p2 = 2q2, which is a contradiction. ` Now suppose p is even. Then q is odd. Now p2 = 2q2, but comparing th 2-adic evaluations yields a contradiction. ` This is a contradiction in both cases, so we’re done.

Theorem 1.5. If

p q n = k ∈ N where m ∈ N, then q = 1.

Proof. Do some blah with prime factorizations.

TODO: Read Chapter 1, Section 1.

1.4 Sets and Classes

Definition 1.6. A set is a collection of elements.

Definition 1.7. A class is a collection of sets. For example, N is a set. Meanwhile, we might consider F as a collection of all finite subsets of N. For example, {1, 6} ∈F . On the other hand, 13 /∈F , while {13} ∈F .

1.5 Constructing the Real Numbers

We will construct R with the method of Dedekin cuts.

Definition 1.8. A cut in Q is a division of Q into two nonempty sets A and B such that (i) A ⊔ B = Q; i.e. A ∪ B = Q and A ∩ B = ∅.

(ii) ∀a ∈ A∀b ∈ B, a < b.

(iii) A has no largest element.

The notation is A | B. Definition 1.9. If B has a smallest element r ∈ Q, we say we have cut the line at r, and this is a rational cut. Example 1.10. Take A = {r ∈ Q : r < 2} and B = Q − A. In this case, min B = 2. So this is a (rational) cut at 2.

Example 1.11. A =

| B = Q − A is not a cut of rational number.

So now we make a definition.

Definition 1.12. A real number is a cut in Q. R is the collection of all cuts in Q.

Now we can impose ordering, etc. on these. We say that A | B ≤ C | D if and only if A ⊆ C. Also, A | B < C | D if and only if A ⊆ C but A̸ = C. Now we can rewrite the least upper bound property as

Theorem 1.13 (Least Upper Bound). If S is a collection of cuts in Q for which S is not empty, and there exists an upper bound of S , then there is a least upper bound for S .

Proof. Take E = {r ∈ QQ : ∃A | B ∈S and r ∈ A} and let F = Q − E. In other words, E = ∪A|B∈S A. We claim E | F is a least upper bound. Now E is not empty because S is not empty. Because A ⊆ E∀A | B ∈S , E is certainly an upper bound. But it must also be a least upper bound, because any other cut misses some element of E, which is contained in some A. Finally we have to check that E | F is indeed a cut. F̸ = ∅ because there exists an upper bound for S , and E does not have a maximal element because none of the A do. Finally, check that all elements of F are greater than all elements of E.

Finally we need to embed Q in our R. This is not automatic, since R is actually a collection of cuts. So, for any c ∈ Q, we associate it with the cut at c, namely

c∗ def = {r ∈ Q | r < c} | {r ∈ Q : r ≥ c} .

This allows us to say Q ⊂ R, even though formally this is wrong. But these identifications allow us to write this anyways. One can also check that c < d ⇔ c∗ < d∗, so the ordering of Q agrees with that of R.

1.6 Arithmetic

Now for the hard part. . . we need to endow R with an operation. Well, addition isn’t too bad

Definition 1.14. For cuts A | B and C | D, we define

(A | B) + (C | D) def = (A + B) | (Q − (A + B)) .

One needs to check a LOT of things. First check that it’s actually a cut, and that all arithmetic properties (commutative, associative, identity . . . ) hold. It’s even worse to define −x given x = A | B. You need to handle the case where B has a smallest element. And so on. Then you need to prove that x + (−x) = 0∗. Now we need to prove equality of sets.

Remark 1.15. The only reliable way to show that A = B is A ⊆ B and B ⊆ A.