One-to-one map of C onto R^1

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In summary: I'm thinking more on the abstract side of things. I would like to arrive at a complex ordering that yields the real ordering as a special case.Ideally (this may be far-fetched), the properties of this complex ordering would be generalizations of order in the real numbers. From the complex orders, the real orders could be derived by simply assuming that Im(z) = Im(q) = 0.
  • #1
JungleJesus
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Can a function be defined such that for a complex argument z = x + iy, the function will uniquely map z onto the real number line? I have a hunch that this would not be possible, but if such a function existed, it could be used to define a unique ordering of the complex numbers without the need for lexicographic ordering, which is not as useful as I would like.
 
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  • #2
Well, C (which as a set is just RxR) and R have the same cardinality, so certainly there is some bijection between them. Do you want it to have some specific properties, like continuity?
 
  • #3
Suppose for a moment that 0 <= x <= 1 and 0 <= y <= 1, and that the map is to the interval [0, 1].

You can interleave the x and y coefficients to get a single number.
For example, if z = 0.1357 + 0.2468i, maps to the real number 0.12345678.
 
  • #4
If I can have a bijection, I would like it to be continuous. I need to be able to use it to define an ordering of the complex numbers that satisfies the properties of order on the real number line (the order axioms).
 
  • #5
JungleJesus said:
If I can have a bijection, I would like it to be continuous.
In other words, you want the line to be homeomorphic to the plane...


I need to be able to use it to define an ordering of the complex numbers that satisfies the properties of order on the real number line (the order axioms).
There's an easy impossibility proof: the ordered field axioms imply that x2>0 for all nonzero x...
 
  • #6
Hurkyl said:
There's an easy impossibility proof: the ordered field axioms imply that x2>0 for all nonzero x...

Sorry about that. I know exponentiation will not preserve order, but the lexicographic ordering that I am familiar with only allows order to be preserved if a constant is added to both sides. If a > b and c > 0, then I would like ac > bc. The lexicographic ordering fails here due to the properties of 0.
 
  • #7
JungleJesus said:
Sorry about that. I know exponentiation will not preserve order, but the lexicographic ordering that I am familiar with only allows order to be preserved if a constant is added to both sides. If a > b and c > 0, then I would like ac > bc. The lexicographic ordering fails here due to the properties of 0.
The ordered field axioms imply that x*x > 0 for any nonzero x.
 
  • #8
Hurkyl said:
The ordered field axioms imply that x*x > 0 for any nonzero x.

Yikes. This is not the way to go. Drop the order axioms. I just want a way to say that any two complex numbers satisfy trichotomy (<, >, =). I was hoping to do this by mapping C onto R via a one-to-one function and then comparing the real numbers produced by that function.

Strictly speaking, if I have z and q (both complex) and G() is my one-to-one (preferably continuous) function of C onto R, then I want to be able to say that:
1.) G(z)=G(q) or
2.) G(z)>G(q) or
3.) G(z)<G(q)

In this fashion, I could define z>q, z<q, and z=q in a similar fashion to the real numbers.

Ideally (this may be far-fetched), the properties of this complex ordering would be generalizations of order in the real numbers. From the complex orders, the real orders could be derived by simply assuming that Im(z) = Im(q) = 0.

As far as that goes, I don't know what's possible and what has already been done.

I do know that ordering the complex numbers is considered to be a null point. I also think, but cannot prove, that any such ordering as the one I described will not be unique.
 
  • #9
What is it about lexicographic ordering that you don't like? It satisfies (if you set it up properly) all the requirements you just gave. If you want some compatibility with the algebraic properties of the complex numbers, then it won't work, as Hurkyl explained.
 
  • #10
Tinyboss said:
What is it about lexicographic ordering that you don't like? It satisfies (if you set it up properly) all the requirements you just gave. If you want some compatibility with the algebraic properties of the complex numbers, then it won't work, as Hurkyl explained.

I'm thinking more on the abstract side of things. I would like to arrive at a complex ordering that yields the real ordering as a special case.

I'm also beginning to look at complex analysis as a major area of study for me. The function G() which I described earlier would surely have some powerful properties. Because G() is one-to-one, it could also offer a unique way to build the complex numbers from the reals. This is big for me. Either system could, in principle, construct the other, given only the definition of G(). The way I see it, that has to mean something...

Also, G() may be able to help derive some complex order axioms which are more general than, say, ac > bc. That's the idea here; there has to be some way of going about this.
 

1. What is a one-to-one map of C onto R^1?

A one-to-one map of C onto R^1 is a function that maps every point in the complex plane (C) to a unique point on the real number line (R^1). This means that for every complex number, there is only one corresponding real number and vice versa.

2. How is a one-to-one map of C onto R^1 different from a regular function?

A one-to-one map of C onto R^1 is a special type of function where each element in the domain (C) is mapped to a unique element in the range (R^1). In a regular function, multiple elements in the domain can be mapped to the same element in the range.

3. What is the significance of a one-to-one map of C onto R^1?

A one-to-one map of C onto R^1 is important in mathematics because it helps us understand the relationship between complex and real numbers. It also allows us to perform operations on complex numbers using real numbers, making complex analysis simpler.

4. Can every complex number be mapped to a unique real number?

Yes, every complex number can be mapped to a unique real number through a one-to-one map of C onto R^1. This is because both complex and real numbers have an infinite number of elements, allowing for a one-to-one correspondence between them.

5. How is a one-to-one map of C onto R^1 related to the concept of injectivity?

In mathematics, injectivity refers to a function that maps distinct elements in the domain to distinct elements in the range. A one-to-one map of C onto R^1 is an example of an injective function, as every complex number is mapped to a unique real number, and vice versa.

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