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### Complex Numbers and Supremum Property

```Date: 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

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
... 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

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,

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,

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.

```

```
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.

interesting problem.

Best regards.

```

```
Thanks very much!

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

Best regards.

```
Associated Topics:
College Imaginary/Complex Numbers

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