Complex Numbers and Supremum PropertyDate: 12/06/2002 at 06:35:56 From: Veronique Subject: Complex numbers and supremum property Hello, I'm a Ph.D. student in commutative algebra working at the Catholic University of Leuven (Katholieke Universiteit Leuven), Belgium. I have also a teaching task, which is helping the students who have problems with the course of analysis. Some time ago, a clever student asked me a question that has puzzled me ever since. It is about complex numbers. I'll denote the field of the complex numbers by C. It is known that one cannot make C into a totally ordered field (since, if i>0, then i*i=-1>0, a contradiction; similarly, the assumption i<0 leads to a contradiction). However, one can define several order relations on C, which are total. E.g. the lexicographic order: a+ib <= c+id if a<c and if (a=c and b<=d). Another example has to do with the geometric representation of complex numbers: let z= |z|exp(ia) and w= |w|exp(ib), with 0 <= a, b < 2Pi. Then z <= w if |z| < |w| or if (|z| = |w| and a <= b). One can check that these relations on C both define a total order. However, for both cases one does not have the supremum property: for both relations; one can easily find a non-empty subset A of C that is bounded above, but that does not have a least upper bound (i.e. a supremum). Now I finally come to my student's question: Is it possible to define a total order on C for which one does have the supremum property? I really don't know the answer. I have already looked in several books, I've asked other mathematicians, I've searched your archives, ... but no one seems to know the answer. The only thing people came up with was a trivial example: if you consider the trivial order relation on C (x <= x for every x in C), then the supremum property holds. But clearly, this is a non-total order relation. So I would be very thankful to receive an answer from you. Thank you for your time. With polite regards, Veronique Van Lierde Date: 12/07/2002 at 18:37:15 From: Doctor Schwa Subject: Re: Complex numbers and supremum property Hi Veronique, Thanks for the fascinating problem! My belief has been that no such total order exists. I like your ideas (lexicographic order has the problem that, say, the set of all complex numbers with negative real part will have no least upper bound; geometric version has a problem with, say, the unit disc including boundary). Now I think I have an idea that DOES give a total order that DOES have the supremum property. Here's my order: For each a+bi, write a and b as their decimal expansions a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0 . a_{-1} a_{-2} ... b_n b_{n-1} b_{n-2} ... b_2 b_1 b_0 . b_{-1} b_{-2} ... where perhaps one of a and b had to be padded with 0s on the left in order to make them the same length. Now map a+bi to the real number c, where c's decimal expansion is a_n b_n a_{n-1} b_{n-1} ... a_1 b_1 a_0 b_0 . a_{-1} b_{-1} a_{-2} b_{-2} ... The usual ordering of real numbers now defines a total order on the c's, and hence on the complex numbers a+bi. Since the c's are real numbers, they have the least upper bound property. Obviously this is far from satisfying the field axioms for any reasonable operations! Does my solution work? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 12/12/2002 at 07:31:03 From: Veronique Subject: Complex numbers and supremum property Doctor Schwa, Thank you very much for your answer. I've checked your solution and I think it does indeed work. With polite regards, Veronique Van Lierde Date: 01/13/2003 at 08:46:39 From: Jacques Willekens Subject: Complex numbers and the supremum property Dear Dr Math, I'm a bit worried by your proposed answer to the question about a total order on C that would have the supremum property. The problem is that the mapping you suggest is not one-to-one. For example, assume that the image in R of a set of complex numbers has 0.0909... as its least upper bound (LUB). If we translate that back to C, we get (0, 0.999...) = (0, 1) = i. Now, if we go back to R, we find that the LUB should also be equal to 1. As the LUB must be unique, don't we get a contradiction ? Of course, we know that C, like every set, can be well-ordered. In such an ordering, if A is a proper subset of C, bounded above, the least element of the set of upper bounds of A is a LUB for A. However, this is not very constructive. Thank you in advance, Date: 01/14/2003 at 20:55:52 From: Doctor Schwa Subject: Re: Complex numbers and the supremum property Hi Jacques, Oooh, that's very good! Thanks for the correction. I think I can dodge your argument by insisting that all the elements of C must be represented in standard decimal form first (that is, insist that elements of C have unique decimal representations, disallowing infinite terminal strings of 9s). Does that salvage my original proposal? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 01/15/2003 at 04:28:54 From: Jacques Subject: Complex numbers and the supremum property Hi Dr. Schwa, Thank you for your quick answer. However, I'm afraid this does not solve the problem. We have, in fact, two possible mappings from C to R: 1) the mapping 'f' (the one you suggest), that uses the standard representation (no infinite strings of 9) 2) the mapping 'g' that uses the alternate representation (no infinite strings of 0; 1 is represented as 0.99...) (We could also use combinations of these, but this is unlikely to simplify matters.) The problem is that, whichever mapping you choose, the inverse mapping from R to C is not properly defined. For example, if we use 'f', the real number 0.0909... does not correspond to any complex number, as the normal candidate (0,0.99..) should be represented by the real number 1 instead. Specifically, if we consider, as in my previous example, the set A = {1/11 - 1/n}, the LUB in R is a = 1/11 = 0.0909... What would be the LUB of the corresponding set in C? If we say that it is (0, 1), we get a contradiction, because, for example, the number (0, 0.1) is less than (0,1) (according to f) and, as it maps to 0.1 in R, it is greater than the elements of A, i.e. it is a smaller upper bound than a. If we try to use the 'g' mapping instead, I believe that we will get the same kind of problem. In this case, we could find an element of the set that is greater than the purported LUB. For example, consider a set B whose LUB in R is 1.1010..., and assume that B contains an element b whose image in R is 0.2323... In this case, the LUB in C should be (1/9, 1), which maps to .1919... in R, and the element b of B is greater than the purported LUB. In fact, what we need is a truly bijective mapping between C and R. Such a mapping exists, because the sets have the same cardinality; however, it is not obvious that that mapping can be specified explicitly. This is something like the Peano curve; I don't know if that kind of mapping is really bijective. There is a similar problem when we attempt to show that the power set of N and the real interval (0,1) have the same cardinality. The way I learned it, we must identify the power set of N with the set of sequences of 0 and 1. If we interpret these sequences as binary numbers, we find that (0,1) <= P(N) i.e., some sequences map to the same real number. If we interpret these sequences as decimal numbers, we find that P(N) <= (0,1) i.e., some real numbers do not map to any sequence. We can then reach the conclusion by using the Schroeder-Bersntein theorem. However, this kind of reasoning cannot be used if we take order into account. I'm not at all sure about this; in any case, it appears to be a very interesting problem. Best regards. Thanks very much! I'm glad our archives have alert readers like you to help me clean up my mistakes. Now I'll have to think about this some more: is there a constructive bijection I can use? If not, is there some constructive ordering on C itself (or on R^2, really) that has the supremum property? Thanks again, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ Date: 01/16/2003 at 03:16:01 From: Jacques To: dr.math@mathforum.org Subject: Complex numbers and the supremum property Dear Dr. Schwa, This was a very interesting discussion; I believe I'll continue myself to think about it. Best regards. |
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