How to define a well-order in C(set of complex numbers)? ord(C)?

This is a question from Set Theory which was on my last exam.





A few definitions to help:





Definition: Partial order in a set A is a relation R in A which is Reflexive, Anti-symmetric, Transitive.





Definition: Let A be an orderd set with the property that every subset of A contains a first element. A is called a well-ordered set.








Well-oredering theorem: Every set can be well orderd.

How to define a well-order in C(set of complex numbers)? ord(C)?
Here's a total ordering for all of C:


Every element of C is of the form a + bi.


for two complex numbers c, d if |c| %26gt; |d| then c %26gt; d, where |a+bi| = a^2 + b^2


If |c| = |d| let c = a + bi, d = f + gi


then when a %26gt; f we let c %26gt; d


If a = f, then either b = g, in which case they are equal, or b = -g, in which case choose the one with the positive part to be bigger.
Reply:This is a total ordering, but it is not a well-ordering. There are many sets with no smallest element in this ordering. For example, {1/n:n in N} Report It

Reply:But first you need an order on C and you can do this lexicographically. So the problem of well ordering C reduces to well ordering the set of real numbers R.





the standard ordering ≤ of the positive real numbers is not a well-ordering, since, for example, the open interval (0, 1) does not contain a least element. There exist proofs depending on the axiom of choice that it is possible to well order the real numbers, but these proofs are non-constructive and no one has yet shown a method to well order the real numbers.(wikipedia)


One way is to follow step by step the well-ordering theorem...


I think what you essentially do is the following:


Say you want to order R( or directly C):


Pick a number x_0 in R and "call" it the least number. Then pick another number x_1 in R\{x_0} which will be the successor of x_1.


You do this method now by transfinite induction. In any step you take out from the remaining set a number which you will "call" the least of the set. Doesn't matter which number you pick. In this way you construct directly not only an ordering but a well ordering. You can pick numbers due to Axiom of Choice. The details remain to you.





You have a well ordering in some abstract way , there are no ways to find a clear way in this case.
Reply:Since the real numbers are well-ordered (because bet ween any two numbers you can put =,%26lt;,%26gt;), first compare the real parts of the complex number. Now the only problem is when the real parts are the same. Well, if the real parts are the same, then you compare the complex parts. This will not contradict the definition of equivalence in complex numbers that two complex numbers are equal if and only if both their real and imaginary parts are equal.





So under this ordering


1+i%26gt;1-i


3i=3i


-3%26gt;-3i
Reply:Unfortunately, there is no way to constructively give such a well ordering. To do so would be equivalent to proving the axiom of choice for sets of cardinality the continuum and that cannot be done in just ZF. That is why AC is assumed. :)