First, on the transfer principle
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The whole point of the transfer principle is that a statement
P is true in the standard model
if and only if *P is true in the nonstandard model.
You can actually prove the "if and only if" if you merely assume the truth of
Q implies the truth of
*Q as follows:
Suppose
*Q is true.
Assume
Q is false, then:
~Q is true.
*(~Q) is true.
~(*Q) is true.
*Q is false.
This a contradiction, therefore
Q is true.
And similarly for the case where
*Q is false.
TMK, for the purposes of NSA, real analysis is cast into the following language:
The language of bounded first-order logic, with:
(1) Constant symbols for every real number and every set that could ever possibly arise in a real-analysis argument.
(2) The internal power set operator
P.
(3) The internal membership relation \epsilon.
"Bounded" means that quantifiers are bound. You can say things like \forall x \in T: x = x, but not things like \forall x: x = x.
"Every real number and set that could possibly arise" means the elements of \mathbb{R}, or of \mathcal{P}(\mathbb{R}), or of \mathcal{P}(\mathbb{R} \cup \mathcal{P}(\mathbb{R})), et cetera.
The "standard model" is simply the following map:
(1) Each constant symbol is mapped to the real number or set it denotes.
(2)
P is mapped to the set-theoretic power set operator \mathcal{P}
(3) \epsilon is mapped to the set-theoretic membership operator \in
Then, the complete first-order theory of real analysis simply consists of every true statement of this language. (If you prefer, it simply takes every true statement as an axiom)
So, Q is true in the standard model
if and only if Q is in this theory.
Every consistent infinite first-order theory has a nonstandard model. (Which means that every statement of the theory must be true in the nonstandard model)
Because our theory of interest is complete, we know that for any statement Q, either Q or ~Q is in the language. It is then easy to show that Q is true in the nonstandard model
if and only if Q is in this theory.
And thus, a statement in this language is true in the standard model
if and only if it is true in the nonstandard model.
Happily, we can find a nonstandard model where:
(1) \mathbb{R} \subset {}^\star \mathbb{R}
(2) \epsilon is mapped to \in
(3) \forall T: P(T) \subseteq \mathcal{P}(T)
The important thing to note, here, is that the internal power-set doesn't contain
every subset -- it only contains the
internal subsets.
Statements, such as Dedekind completeness, can be made within this first-order language. (Using the internal power-set function, and the internal membership operation)
Since Dedekind completeness is true in the standard model, it must also be true in the nonstandard model. (According to the aforementioned interpretation)
The resolution to the apparent paradox is that Dedekind completeness says the following:
Every
internal Dedekind cut of
R is generated by an element of
R. (Why internal? Because we formulate it using the internal power-set operator)
When we pass to the nonstandard model, the above statement must be true. Since the converse is also true, we have the following external theorem:
A Dedekind cut of *
R is internal
if and only if it is generated by an element of *
R.
Hopefully this exposition resolves your issues about completeness and the transfer principle!
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We can, of course, study the ordered field *
R without doing it the nonstandard way.
*
R is ordered, so it has a topology. Its open sets are simply those of the form (a, b), where a and b are hyperreal numbers. For example, this means the definition of the limit of a function is entirely unchanged! (It's still given by the epsilon-delta formulation)
Among *
R's topological properties is:
(1) It's totally disconnected. (Yuck)
(2) Every convergent countable sequence is eventually constant.
(3) Every countable subset of *
R is bounded.
*
Z is an
integer part of *
R. This means that it is a discrete ring with the property that for every element r of *
R, there exists an n in *
Z such that n \leq r < n+1.
*
N is, of course, the nonnegative subset of *
Z.
We have the following theorem:
Let
F be any ordered field.
Let
I be an integer part of
F.
Define an I^+-indexed sequence by: a_n := 1/n.
Then, \lim_{n \rightarrow +\infty} a_n = 0.
Recall that, by definition, this means:
<br />
\forall \epsilon \in F^+ : \exists N \in I^+: \forall n \in I^+: (n > N) \implies (|a_n| < \epsilon)<br />
Which is easy to prove, by the definition of "integer part". (There exists an
M such that M \leq 1/\epsilon < M+1)
There are, of course, many different nonstandard models of
R, and they have different external properties. For example, you can ask things like "What is the minimum length for an unbounded sequence?" or "What is the cardinality of *
R?"
There are other interesting properties too. If I recall correctly, there exist
Cauchy complete nonstandard models of
R. However, a Cauchy sequence cannot be of countable length -- it must have an uncountable index set.
There's a refinement of Dedekind completeness that means the same thing, I think: the separation between the two parts of the Dedekind cut is not allowed to be bounded below by a positive number.
So, for example, the Dedekind cut:
(infinitessimals, and everything negative) : (positive finites and infinites)
is not fair game, because the gap between them is "too big".
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These ideas extend into the algebraic domain in two ways.
The "baby version" of all of this is the theory of real closed fields -- it only concerns itself with the semi-arithmetic operators (+, *, <, etc), and doesn't try to model any set theory.
A good example of a real closed field is the set of real algebraic numbers.
A version of Dedekind completeness is valid for any real closed fields. (E.G. if you are only using the semi-algebraic operations, you'll never know you're missing pi)
Similarly, if you are willing to pretend that only semi-algebraic sets exist (things definable in terms of polynomial equations and inequalities), any real closed field is actually fairly nice, topologically. (This, I believe, is closely related to "internal sets" in nonstandard analysis, and is one of the issues that drives my current interests)
The other connection is that algebraists have defined something they call a
hyperreal field, whose definition you can see at
Wikipedia which are related to nonstandard models of the reals. (some of the hyperreal fields are some of the nonstandard models, and I suspect they
all are, from the right perspective)
Ack, I'm rambling. I'll stop now!
, I am glad you think me so skilled but I am far from it. I am only 3 years older than your 17 by a few months and am a first year maths student in University. In fact, I too did not learn any mathematics in high school (no geometry or calculus, only algebra). I reaped the public school system to its full benefit and did not learn much math through it. But at your age, 17, I decided to buckle down and teach my self a bunch of stuff since I am interested in fundamental questions. Between 17 and 19 i changed degrees a few times - from Computer science (1/2 a year) to computer engineering (1.5 years) before now settling with math. But yeah almost all my knowledge is self gained and taught. I am soon to begin study of algebraic topology...
