14.Consistency and Independence

The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Quote attributed to the mathematician Jerry Bona

Most mathematicians would probably regard the axiom of choice as 'obviously true,' while others may regard it as a somewhat questionable assertion which might even be false( and I am myself inclined towards this second viewpoint). Still others would take it as an assertion whose 'truth' is a matter of opinion or rather, as something which can be taken one way or the other, depending upon which system of axioms and rules of proceedure one chooses to adhere to.

The Road to Reality

1931 -

The issues

The Up Side

Zorn's "Template"

Let $\QTR{Large}{S}$ be a Set. Suppose we have defined a "structure" on $\QTR{Large}{S}$ .(I will not tell you what a structure is). Next, suppose we have a proposition $\QTR{Large}{\Phi }$ defined on . $\QTR{Large}{\Phi }$ is stated in terms of the structure on $\QTR{Large}{S}$ .

Restrict attention to , $\QTR{Large}{P}$ being the subsets such that

Order $\QTR{Large}{P}$ by inclusion of subsets.
Prove that if is a chain in $\QTR{Large}{P}$ then
Conclude that there is a maximal element $\QTR{Large}{\in P}$ . That is

Subjective Concerns

There are discontinuous additive functions -

Discontinuous functions which satisfy the equation:

Proof:

Consider as a vector space over . Extend $\QTR{Large}{\{}$ 1 $\QTR{Large}{\}}$ to a basis for . We will write it as $\QTR{Large}{\{}$ 1. Define a map

1

by setting $\QTR{Large}{f(}$ 10 and 1 for all (anything works as long as one 0). Extend $\QTR{Large}{f}$ to all of , as a map of vector spaces (hence it is additive).But, 0 for all . Yet 0 for all . This contradicts 19.5.
There are non-measurable Sets -

A definition of Lebesgue outer measure

A definition of Lebesgue measure

Non-measureable Sets
The Banach-Tarski Theorem(Paradox) - It is possible to decompose the 3-sphere " into a finite number of pieces and then reassemble them into two identical 3-spheres using only rigid motions.

Practical Concerns

Well Orderings of the Reals - Where are they? What is the value of an existence proof?

Consistency and Independence

We begin by reviewing the concepts of consistency and independence of axioms in the context of of Peano's Postulates for arithmetic. We begin with a less-than formal definition of a proof for a system of axioms.

14.1 Very Loose Definitions:

Given a list of axioms A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,......A $_{\QTR{Large}{n}}$

A "proof" of S $_{\QTR{Large}{k}}$ is series of statements and symbols of the form S $_{\QTR{Large}{1}}$ ,S $_{\QTR{Large}{2}}$ ,S $_{\QTR{Large}{3}}$ ,......,S S $_{\QTR{Large}{k}}$ where each S $_{\QTR{Large}{i}}$ is either an axiom or follows from S $_{\QTR{Large}{1}}$ ,S $_{\QTR{Large}{2}}$ ,S $_{\QTR{Large}{3}}$ ,......,S by a rule of inference. We also say S $_{\QTR{Large}{k}}$ can be proved from A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,......A $_{\QTR{Large}{n}}$ .
An example of a rule of inference is Modus Pones P ,PQ Q . If P is true and P impliesQ is true, then $\$ Q is true.
A list of axioms is said to be consistent if there is there is no statement S for which S and S (not S) can be proved.
A statement S is independent of axioms A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,......A $_{\QTR{Large}{n}}$ if neither S nor S can be proved from the axioms. In particular, that the axiom lists A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,......A $_{\QTR{Large}{n}}$ ,S and A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,......A $_{\QTR{Large}{n}}$ ,S are consistent.

14.2 Another Loose Definition:

A mathematical model for a list of axioms is a well-defined set which assigns "meaning" for the undefined terms in the axioms, in a manner such that the axioms are "true". The existence of a model proves the consistency of a system since one "knows" that a statement and its negation cannot both be true in a model.

14.3 An Example:

A set $\QTR{Large}{S}$ , a distinguished element 1 $\QTR{Large}{\in S}$ , a a map is said to satisfy Peano's Postulates if

A $_{\QTR{Large}{1}}$ For all

A $_{\QTR{Large}{2}}$ For all

A $_{\QTR{Large}{3}}$ For all C

A $_{\QTR{Large}{4}}$ If is a proposition such that:

1 .
For all ,

Then For all ,

, 1 is a model for Peano's Postulates. In fact, it is really the only model up to an appropriate notion of equivalence.

Note that 2,4,6,.. $\QTR{Large}{\}}$ "1" $\equiv$ 2 and 2 also satifies the Postulates

However, Consider the list A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,A $_{\QTR{Large}{5}}$ . A $_{\QTR{Large}{4}}$ this is consistent since is a model .

Thus we can conclude that A $_{\QTR{Large}{4}}$ is independent of A $_{\QTR{Large}{1}}$ ,A $_{\QTR{Large}{2}}$ ,A $_{\QTR{Large}{3}}$ ,A $_{\QTR{Large}{5}}$ .

What is the relationship between The Zermelo-Fraenkel Axioms, The Axiom of Choice, and The Continuum Hypothesis?

The Zermelo-Fraenkel Axioms(ZF):

The Axiom of Choice(AC):

Let $\QTR{Large}{S}$ be a set and . There exists at least one function $\QTR{Large}{f}$ ,

such that $\$ for each set

The Continuum Hypothesis(CH):

There is no set $\QTR{Large}{S}$ such that

The Answer - They are independent of each other!

In 1935 Kurt Gödel showed that both AC and CH are consistent with ZF, in that neither AC nor CH can be proved from ZF. In particular he discovered a model of ZF in which both AC and CH are true.

In 1963, Paul Cohen proved independence AC and CH by first constructing a model of ZF+AC in which CH fails, and then a model of ZF+CH in which AC fails. Together with Gödel's models, this demonstrates that neither the Axiom of Choice nor its negation can can be proved from The Zemelo-Fraenkel Axioms, and that neither the Continuum Hypothesis nor its negation can be proved from Zemelo-Fraenkel Axioms, even in the presence of The Axiom of Choice.