0. Background

What is a Theory?

In Mathematics, a Theory is a collection axioms and the collection of all statements that can be derived from them via logical inference.

A theory provides "concrete" model for a "mathematical concept." For example, Metric Spaces provide a model for sets with a concept of distance between members of the set.

What is a Theorem?

A Theorem is a statement that can be derived from the axioms of a Theory.

What was Eilenberg telling us?

Once we know the axioms, the theorems are determined. Of course, since proofs may be very hard to come by, finding the theorems for a given theory may be difficult.

In the other hand, the question as to whether a given theory provides a good model may not have a simple true or false answer. In particular, if a theory does not provide a good model, the theorems may be true but may not be particularly interesting or useful.

Indeed, this question, regarding Set Theory will occupy our attention for much of the Semester

Objects:

Here is a list of familiar objects that will be of interest to us this semester.

The Natural numbers {1,2,3,4,.....}
The Integers {.....,-3,-2,-1,0,1,2,3,4,.....}
The Rational numbers
{ a $\QTR{Large}{=}$ 0, b $\QTR{Large}{=}$ 1 or pairs a,b with a0 $\QTR{Large}{\}}$ b and gcd( $\mid$ a $\mid$ ,b) $\QTR{Large}{=\ }$ 1)}

The reason for all this formality is to be sure that there is a unique representation of each rational.
The Real numbers
$\QTR{Large}{(}$ 0,1 $\QTR{Large}{)}$ All Real numbers x such that 0 $\QTR{bs}{<}$ x $\QTR{bs}{<}$ 1. More formally, {x $\QTR{Large}{|}$ x and 0 $\QTR{bs}{<}$ x $\QTR{bs}{<}$ 1}
The Complex numbers
Many Others

Language:

First some logical notation:

The symbol is to be read "there exists."
The symbol is to be read "for every."
The symbol is to be read "not"
The symbol is to be read "implies"

The context that this notation will appear will be in the "language of Sets." In particular, whether or not a given property holds will depend on the Set that it belongs to.

Here are two examples, the second of which we will write in the more informal notation that we will use almost everywhere in these notes :

The first example, in increasingly formal language, is a statement that is true for and , but not for . Try to solve 3z $\QTR{Large}{=}$ 2.
- For all x and y, if x $\QTR{Large}{\neq }$ 0 then there exists a z such that xz $\QTR{Large}{=}$ y.
- x $\QTR{bs}{\forall }$ y $\QTR{Large}{(}$ if x $\QTR{Large}{\neq }$ 0 then $\QTR{bs}{\exists }$ z $\QTR{Large}{(}$ xz $\QTR{Large}{=}$ y $\QTR{Large}{))}$
- x $\QTR{bs}{\forall }$ y x $\QTR{Large}{=}$ 0 $\QTR{Large}{)}$ $\QTR{bs}{\exists }$ z $\QTR{Large}{(}$ xz $\QTR{Large}{=}$ y
Let f:[0,1] $\rightarrow$ [0,1] be a continuous function then there exists an x such that f(x)=x. You might remember the proof is just the intermediate value theorem, using g(x)=x-f(x). One notes that g(0) $\leq$ 0 and g(1) $\geq$ 0.The statement is not true if you replace [0,1] by . Just choose f(x)=x+1.

$\vspace{1pt}$

$\vspace{1pt}$ The Algebra of Sets

Sections 1.1, 1.2, 1.4, and 1.5 cover what should be fairly familar territory in elementary set theory, unions, intersections, functions, equivalence relations, etc. (Assignment: Read this material and bring any questions to the next class)

We will want to look at products of sets. That is if $\QTR{Large}{A}$ and $\QTR{Large}{B}$ are sets and

eg: The set of ordered pairs of Natural Numbers.

If $\QTR{Large}{A}$ and $\QTR{Large}{B}$ are sets, we will want to look a sets of functions (we will often use the term "map") from $\QTR{Large}{A}$ to $\QTR{Large}{B.}$ We use the notation

There is some important algebra here as well.

0.1 Definitions:

a. Fix $\QTR{Large}{A}$ . Suppose we are give a map . We define a map by the formula for

b. Now fix $\QTR{Large}{B.}$ Suppose we are give a map . We define a map by the formula for.

0.2 Lemma;

In the setting of 0.1

a. Suppose is one to one then is one to one.

b. Suppose is onto then is one to one.

Proof: Exercise(hint: immediate from the definitions)

0.3 Lemma:

a. Let and be sets. Let and be the inclusions. Then, for any set $\QTR{Large}{B}$ , the map

is 1 to 1. If and are disjoint then it is also onto.

Note: , the ordered pair of maps $\QTR{Large}{f}$ restricted to $\QTR{Large}{A_{1}}$ and $\QTR{Large}{f}$ restricted to $\QTR{Large}{A_{2}}$ .

b. Let and be sets. Let be the projection onto the "ith" axis. For any set $\QTR{Large}{A}$ , the map

is a 1 to 1 and onto.

Proof:

a. Cutting through the notation, the proof is little more than observing that any map Is uniquely determined by its values on $\QTR{Large}{A_{1}}$ and $\QTR{Large}{A_{2}}$ .

If and are disjoint then any pair of maps can be combined to give a map on . If not, only pairs

agreeing on can be so combined, hence is only 1 to 1.

b. is even more straight forward.