Mon, 08 Feb 2016 08:01:46 GMT

This was again a presentation for the
Best mathematics club in the multiverse. It is meant for both
younger and older students, as well as working mathematicians. Its
objective therefore lied in breadth rather than depth. I removed lots
of topics I wanted to talk about, as the talk I gave got too
long. **It is not a scientific talk.** Comments and corrections are
welcome.

Why are well-orderings the backbone of mathematics? A similar claim for Zermelo-Fraenckel-Set-Theory (ZFC) was made by Oliver Deiser in his book Einführung in die Mengenlehre, about the constructible cardinal numbers. In ZFC, cardinal numbers are specific ordinal numbers, and ordinal numbers are sets which are well-ordered by the -Relation (roughly speaking). And in fact, the study of well-orderings is an important part of set theory. In recursion theory, well-orderings and well-quasi-orderings can be used for termination proofs, and are also important for newer foundational systems that base on type theory. In proof theory, well-orderings can measure the consistency strength of a theory. The axiom of choice is closely related to well-orderings, and it is an important principal in classical mathematics, specifically analysis, stochastics and topology.

**Definition.** *In short, a well-ordering is a linear ordering
relation with no infinite descending chains.*

That is, if we have some collection of objects ("Set") , and a binary relation on this collection, then the pair is a well-ordering, if it satisfies:

**linearity**: .**irreflexivity**:**transitivity**:**well-foundedness**: Let and . Then has a -minimal element.

We write for , and for and for . In most cases it is clear from the context which relation we mean, so we shall leave the index out and write instead of , etc. We often just write instead of if the ordering we mean is clear (or no specific order is given).

Canonical examples for well-orderings are the default ordering on , and on all its subsets – especially also all of its finite subsets, and the empty set (no axiom sais that there has to be an element at all).

**Exercise:** Show that if is a well-ordering, then
is reflexive, transitive, antisymmetric
() and
every decreasing sequence is stationary.

**Definition:** We call orderings which are reflexive, transitive,
antisymmetric, and in which every decreasing sequence is stationary,
**weak well-orderings**.

Most people have probably heard of the concepts of homomorphism and isomorphosm. The idea is that, given two structures, there is a "renaming" between these structures that preserves the structure fully or partially.

**Definition:** Let and be well-orderings, and
, such that . Then is called a
*homomorphism*. A bijective homomorphism is called *isomorphism*.

(Notice that by definition, every homomorphism is injective.)

Let us define , with the canonical ordering on it.

**Lemma:** For all , if and
only if there exists a homomorphism .

*Proof.* "": Just set . "": This is an instance of the well-known
Pigeonhole principle.

**Lemma:** Every finite subset with
elements is isomorphic to .

*Proof.* By induction. For this follows from
extensionality. As step, notice that is non-empty, and
therefore has a minimal element, say . By induction, there is
an isomorphism . Then
trivially, is an isomorphism.

**Lemma:** Every infinite subset is
isomorphic to .

*Proof.* Let us define

From our previous lemma, we get isomorphisms . Notice that the graphs of these isomorphisms are strictly increasing, that is, . Therefore, we can define , and this must be the graph of a bijective function .

**Lemma:** Induction is equivalent to well-ordering on
.

*Proof.* "": Let , and
. Then there is some . We do
induction on . For , is trivially
minimal. Now in the step, if we already know that all sets containing
a number have a minimum, and , either
is already the minimum of , or contains a
smaller element. "": Assume there was some set
such that , but
. Then there is some minimal element in
. Trivially, and
. Contradiction.

Side Note: This lemma is related to **Markov's Principle**, which is a
theorem about intuitionistic logic. It is an example for a theorem
which holds for every instance, but is not provable in general.

**Exercise:** If this is new to you: Before you read on, can you think
of infinite well-orderings which are **not** isomorphic to
?

*Solution.* Of course, there is no unique solution to this. One simple
example is the canonical ordering on . Essentially, it is almost the
same as , with one exception: is larger than
all other elements.

Something very common in mathematics is to view structures "up to
isomorphism". For example, one talks about the "Field ",
even though in the common mathematical theories it is not unique. We
will do the same with well-orderings now: We *identify* isomorphic
well-orderings. Therefore, we can say, all finite well-orderings are
given by . We already began comparing finite
well-orderings by their homomorphisms, and we will just continue doing
so by saying, a well-ordering is less-than-or-equal a
well-ordering , written , if there is a
homomorphism . Then, we go even further, and
leave out the squared brackets, that is, we write
for the finite well-orderings.

We have already seen that the smallest infinite well-ordering is the canonical ordering on , which we will call .

**Lemma:** Let be a well-ordering, and . Then is also a well-ordering
which is not isomorphic to .

*Proof.* Trivially, is also a
well-ordering. Assuming there was an isomorphism , then is
non-empty and contains a smallest element such that
. Then there must be some such that
, but as was minimal and , we
know that , therefore , so ,
which means that since was minimal, and
therefore . Contradiction.

**Lemma and Definition:** For every two well-orderings , the
sum is defined as the disjoint union of and
with the ordering that makes every element of
smaller than every element of . is a
well-ordering.

*Proof.* Exercise.

**Notice:** The addition of well-orderings is not
commutative. , while is the
well-ordering in our above example.

**Lemma and Definition:** For every two well-orderings , the
product is the lexicographical ordering on
: .

**Notice:** The product of well-orderings is not
commutative. , but .

**Notice:** The finite well-orderings with addition and multiplication
are essentially .

While currently, in the realm of logic, there are new tools being built to cope with this problem, for well-orderings the problem has been solved in an elegant way inside set theory, which is currently the one which is most well developed.

**Definition:** A set is called an **ordinal number** if for
all we have and is
well-ordered by . The class of all ordinals is called
.

**Lemma:** for .

*Proof.* contains a minimum y. Assume there was some , then also , and would not be minimal
anymore. Contradiction. Thus, .

**Lemma:** If and , then .

*Proof.* As , also , so trivially,
is well-ordered by . Assuming and
, by transitivity we have , so every
element of is a subset. .

**Theorem:** If and are ordinals whose
well-orderings regarding are isomorphic, then .

*Proof.* Call the isomorphism . Assuming ,
there must be a minimal such that . Trivially, , so
. Therefore, there is some . Trivially, we have . Contradiction.

**Lemma:** If , then
.

*Proof.* By the foundation axiom, we get an such that
. Take any . ,
because . By linearity of
therefore .

**Lemma:** If is a set of ordinals, then is
an ordinal.

*Proof.* We first prove that is an ordinal number. If
, then , therefore
, therefore . We still have to prove that is a well-ordering on
. Irreflexivity follows from the foundation axiom: No
set may contain itself. For linearity, consider , that is, for all . Since
is an ordinal, linearity holds for
. Well-foundedness of follows from the
foundation axiom. Transitivity is left as an exercise.

Chaining a few more technical proofs, we could prove:

**Theorem:** For any two , we have .

**Corollary:** Every set of ordinals is well-ordered by the relation.

We can generalize induction on to induction on :

**Theorem (Transfinite Induction):** If for some proposition
we have , and , then .

*Proof.* Assume not. Then there is some with , and therefore is a
well-orderet set and has a minimum , which is also the
absolute minimum value with . Trivially,
, and by its minimality, also , but this would imply
. Contradiction.

**Lemma:** is not a set.

*Proof.* If was a set, . Contradiction.

**Lemma:** Every well-ordering is isomorphic to some
ordinal number.

*Proof Sketch.* Define . Now let be
defined for all . If there is still an remaining, set . Otherwise, proceed. Since we
assumed that is a set, this procedure must terminate at some
point.

The nice thing is that in this theory we have concrete objects to talk about, namely the ordinal numbers.

Zorn's Lemma is one of the three classically equivalent axioms independent of ZF: The well-ordering theorem, which says that every set has some well-ordering; the axiom of choice; and Zorn's Lemma. It is probably the one that is used most often explicitly outside of set theory.

**Zorn's Lemma:** If is transitive, reflexive and every
non-empty chain has an upper bound, then contains a maximal
element.

**Notice:** A maximal element might not be comparable with many other
elements.

**Definition:** A **Well-quasi-ordering** is a **well-ordering**
except for it may be nonlinear.

In some sense this generalizes the notion of trees. Zorn's Lemma can be reduced to some stronger property about well-quasi-orders:

**Theorem:** If is transitive, reflexive and every
non-empty well-ordered subset has an upper bound, then for all
, contains a maximal element which is
comparable to .

This essentially sais that if every branch of the tree is bounded, then there are leaves.

A vector space over a field is defined by a few axioms that can easily be looked up in wikipedia.

**Theorem:** Every vector space has a base.

*Proof 1.* Take an arbitrary element. If the hull of the element is
the whole space, you are done. Otherwise, take an element which is
not in the hull and proceed. As the vector space is a set, this
process must "end" after infinitely many steps: After "set" many
steps.

*Proof 2.* All chains of bases are bounded. Therefore, by ZL, there is
a basis.

Proof 1 is normally only done in the finite case, but also works in the infinite case, because of transfinite induction. Proof 2 uses Zorn's Lemma. Similarily, algebraic closure and many other theorems use Zorn's Lemma.

**Fun Fact:** The groups and are isomorphic.

*Proof.* It is easy to show that both are vector spaces over
, and as such, they have a countable base.

Consider a simple programming language: You have an arbitrary number
of registers `a[i]`

which are initially `0`

and can store natural
numbers. You have a command `a[i]++`

which increments the register
`a[i]`

, and `a[i]--`

which decrements it if it is not `0`

. (We do not
allow constructs like `a[a[i]]`

, the `i`

must be constants.) You have
an `exit`

command. Furthermore, you have a command ```
if a[i] == 0 then
goto l1
```

where `l1`

is the line number to jump to. It is easy to show
that this language is turing complete. So especially, it is generally
undecidable, whether a program terminates.

However, we can change the semantics a bit: Attach every program line
with an additional function from ordinal numbers into ordinal
numbers (or, if you want to implement it on a real computer, other
well-orderings), which must satisfy the condition that it is strictly
sublinear, that is, for all
. In the beginning, you start with *some* ordinal
number, that will, at every step, be decreased through the
corresponding function. The program terminates on `exit`

as before, or
as soon as the ordinal is .

All programs in such a language terminate. What programs can be expressed depends on the values of the original ordinal number that we allow. By the way, this is not pure theory.

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