# Analysis axioms?

1 view

### G. A. Edgar

Oct 11, 1995, 3:00:00 AM10/11/95
to
In article <45gtrf\$k8\$1...@mhafc.production.compuserve.com>, Gourio
<10066...@CompuServe.COM> wrote:

> Has everyone heard of axioms used for (real) analysis? Is there a
> construction of this part of mathematics? Which should be these
> axioms? I've read Peano system for natural numbers and of Hilbert
> geometry construction, but is analysis too hard to be put in
> axioms? And is there a problem with infinite?

There are axioms to characterize the real number system: in words,
complete ordered field ("complete" understood in the sense of order).

That, together with the axioms for ZFC set theory are enough to
do real analysis. (ha, ha; a joke...)
--
Gerald A. Edgar ed...@math.ohio-state.edu
Department of Mathematics
The Ohio State University
Columbus, OH 43210

### Dave Seaman

Oct 11, 1995, 3:00:00 AM10/11/95
to
In article <45gtrf\$k8\$1...@mhafc.production.compuserve.com>,
Gourio <10066...@CompuServe.COM> wrote:
>Has everyone heard of axioms used for (real) analysis? Is there a
>construction of this part of mathematics? Which should be these
>axioms?

The axioms for the real numbers can be very concisely stated: the real
numbers form a complete ordered field.

The "ordered field" part is a purely algebraic concept, and you are not
likely to find a thorough explanation of this in an analysis text. You
need to look at a text that explains the concept of a field, such as
Herstein's _Topics_in_Algebra_. The rational numbers are an example of
an ordered field.

The completeness axiom is the one that is mainly of interest in
analysis. It is equivalent to the least upper bound axiom: any set of
real numbers that is bounded above has a least upper bound. Notice
that the corresponding statement for rational numbers is false.

Dave Seaman

### Gourio

Oct 11, 1995, 3:00:00 AM10/11/95
to
Has everyone heard of axioms used for (real) analysis? Is there a
construction of this part of mathematics? Which should be these
axioms? I've read Peano system for natural numbers and of Hilbert
geometry construction, but is analysis too hard to be put in
axioms? And is there a problem with infinite?
François

### Toby Bartels

Oct 14, 1995, 3:00:00 AM10/14/95
to
Gourio <10066...@CompuServe.COM> writes:

Here are the axioms for a complete ordered field:

An operation # on a set R is a function from R x R -> R
(# takes any two numbers in R and returns another number in R).
Defn: (x # y) = x # y is the number to which # maps x and y.

An abelian group is a set R with an operation # such that:
There exists an identity e such that, for all x in R, e # x = x = x # e.
For all x in R, there exists an inverse x' such that x # x' = e = x' # x.
Always, (x # y) # z = x # (y # z) -- associativity.
Always, x # y = y # x -- commutativity.

A field is a set R with operations + and * such that:
(R, +) is an abelian group.
Defn: 0 is the identity of +.
Defn: - x is the inverse of x under +.
(R \ {0}, *) is an abelian group.
Defn: 1 is the identity of *.
Defn: x^{- 1} is the inverse of x under *.
(The only reason (R, *) is not an abelian group is the nonexistence of 0^{-1}.)

An ordering < on a set R is a relation between R and R
(that is, for any x and y in R, x < y is either true or false)
such that:
Always, x < x is false -- antireflexivity.
If x != y, either x < y or y < x but not both -- antisymmetry.
(These may be combined as:
exactly one of (x < y, x = y, y < x) is true -- trichotomy.)
If x < y and y < z, then x < z -- transitivity.
Defn: x is a lower bound of the subset A iff, for every y in A, y < x is false.
Defn: x is an upper bound of the subset A iff, for every y in A, x < y is false.
Defn: x is a least member of the subset A iff x is in A and x is a lower bound.

A complete ordering is an ordering such that:
For every subset A except the empty set, if A' is the set of upper bounds of A,
then A' has a least member unless A' is the empty set.
(In short, every nonempty subset with an upper bound has a least upper bound.)

An ordered field is a field with an ordering such that:
If x < y, then x + z < y + z.
If x < y and 0 < z, then x * z < y * z.

A complete ordered field is an ordered field whose ordering is complete.

Theorem: all complete ordered fields are isomorphic,
with respect both to the ordering and to both operations.
(Proof omitted.)

Thus, the real numbers are referred to as `the complete ordered field'.
^^^

-- Toby
to...@ugcs.caltech.edu

### Mark Hopkins

Oct 16, 1995, 3:00:00 AM10/16/95
to
The axioms are these:

FIELD AXIOMS:
0, 1 are distinct elements
a + 0 = a = 0 + a, a1 = a = 1a
a + b = b + a, ab = ba
(a + b) + c = a + (b + c), (ab)c = a(bc)
for every a, there is a -a such that a + -a = 0
for every a except 0, there is a a' such that a a' = 1
a (b + c) d = abd + acd

ORDERING AXIOMS:
a not < a; a < b --> b not < a;
a < b & b < c --> a < c
a not < b --> b < a or b = a

addition is monotonic: a < b --> a + c < b + c
positive multiplication is monotonic: 0 < a < b, 0 < c --> ac < bc

COMPLETENESS AXIOM:
This property, which states that every possible gap is filled, cannot be
expressed in any recursively defined set of axioms, but can be approximated by
a set-theoretic or first order axiom scheme both of which have the intent of
sorta quantifying over all possible properties of real numbers (and thus all
possible gaps). A set-theoretic version would read:

If A and B are subsets of the real numbers R such that:
(i) A, B are convex sets
(ii) A union B = R
then
one of the sets has a point which lies arbitrarily close to
both sets.

That point represents the "gap" between the sets A and B which is "filled in".

There is a two dimensional version of this same property that could be
used to directly axiomatize R x R:

If A, B and C are subsets of R x R such that:
(i) A, B and C are all convex sets
(ii) A union B union C = R x R

then one of the sets contains a point that lies aribitrarily close to points
of all three sets.

A convex set S in 1-dimension would be one with the property

if a, b are in S,
then for all x between a and b (i.e. a < x < b), x is in S.

In two dimensions, "between" would replace the "a < b" relation, and you'd
have to formulate axioms for betweenness similar to the ordering axioms.
Some, analogous to those above, would take the form:

a is not between a and anything else
a between b and c --> b not between a and c
a between b and x, b between c and x --> a between c and x

The axioms for analysis in multiple dimensions then are one and the same
as the axioms for Euclidean geometry in that dimension -- plus the
completeness axiom.

So analysis and geometry are actually the same thing. So ... the brief

Analysis Axioms = Euclidean Geometry Axioms + Completeness Axiom

The reason the complete + ordered + field axioms cannot precisely define
the reals is because you can always force an unintended interpretation simply
by adding an infinite set of axioms of the forms:

0 < W, 1 < W, 2 < W, 3 < W, 4 < W, ...

where W is a new formal symbol. Since all theorems and proofs are finite
they only involve a finite number of axioms -- in particular a finite number
of these axioms -- but any finite subset of these axioms are consistent,
therefore no contradiction can ever be derived from adding the entire set of
axioms (unless real numbers don't exist at all!) However, the only possible
interpretation for W is that it is an infinitely huge number with an infinitely
wide decimal expansion.

So that's the archetypical formulation of a Non-Standard model of the
analysis axioms and leads to a form of Non-Standard Analysis. To actually
prove the existence of a non-standard model (or, for that matter, the
relative consistency of the real numbers) you need to assume the existence
of an ultrafilter, or something equivalent in power. But that's another
story.

### theodore hwa

Oct 17, 1995, 3:00:00 AM10/17/95
to
Mark Hopkins (ma...@omnifest.uwm.edu) wrote:
: The axioms are these:
: [field axioms,,
ordering axioms,
: and completeness axiom.]

: The reason the complete + ordered + field axioms cannot precisely define

: the reals is because you can always force an unintended interpretation simply
: by adding an infinite set of axioms of the forms:

: 0 < W, 1 < W, 2 < W, 3 < W, 4 < W, ...

: where W is a new formal symbol.

What is W a symbol for? Is it a symbol for a real number? I.E., are the
axioms expressing the statement "there exists a real number W such that
W > any integer ? " (Yes, I know there's some sort of hierarchy of
first-order, second-order, etc. axioms, and my statement is on a higher level
than any of the individual axioms you listed above [right? -- I'm not
completely sure], but I just want to make sure I understand what you mean when
you list those axioms above.)

Since all theorems and proofs are finite
: they only involve a finite number of axioms -- in particular a finite number
: of these axioms -- but any finite subset of these axioms are consistent,
: therefore no contradiction can ever be derived from adding the entire set of
: axioms (unless real numbers don't exist at all!)

However, the only possible
: interpretation for W is that it is an infinitely huge number with an infinitely
: wide decimal expansion.

I see how all this is true, and the last sentence here suggests that W
is intended to represent another real number. Looking at the consequences
of doing this leads to some interesting things. (I'm thinking of all this
on the fly and typing it, so perhaps some places my descriptions could be
patched up / reworded a bit.)

First thing is, what is W + 1? It certainly cannot be any real number
< W nor W itself without violating the other axioms. Now if you say,
let's throw in all real (*) linear combinations of powers of W (**), so that
it's closed under the field operations, how would you compare two such linear
combinations? (Let me call these real linear combinations of powers of W,
"polynomials in W".) Since it is enough to specify which of these
are positive, how would you tell if such a thing is a positive? One way
is to say that a polynomial in W is positive if its leading coefficient
is positive, and the quotient of two such polynoimals positive, if both
are positive or both are not. This satisfies all necessary ordering axioms
and the only question is, is it complete? I think so although I can't come
up with a proof right now.

(*) that is, real numbers between -W and W
(**) and their quotients, excluding zero denominators, and defining the usual
equivalence relation a/b = c/d iff ad = bc (i.e., take the quotient
field)

Now I think this example is of a complete(?), ordered field not isomorphic
to R is easier to understand than the usual one involving Laurent series
with finitely many terms having negative powers of x. What do you think
of this and everything else I've said?

But this field is not Archimedean (W is bigger than every integer). So
do the axioms Archimedean + Completeness + Ordering + Field axioms determine
the reals? I remember this being a standard result, right?

### Dave Seaman

Oct 17, 1995, 3:00:00 AM10/17/95
to
>Mark Hopkins (ma...@omnifest.uwm.edu) wrote:

>: The reason the complete + ordered + field axioms cannot precisely define
>: the reals is because you can always force an unintended interpretation simply
>: by adding an infinite set of axioms of the forms:
>
>: 0 < W, 1 < W, 2 < W, 3 < W, 4 < W, ...
>
>: where W is a new formal symbol.

But then the least upper bound axiom fails. Consider the integers.
This set is nonempty, and it is bounded above (by W). What is the
least upper bound?

There is none, because 0.1*W is an upper bound, and so is 0.01*W, and
0.001*W, and ....

Therefore, you don't have a complete ordered field.

Dave Seaman