Are Real Numbers R a Subset of Complex Numbers C?

[johann1301]

Are the Real numbers R a subset of the complex numbers C?

 

[Greg Bernhardt]

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

 

[homeomorphic]

Yes, they are a subset, unless you are really pedantic about it.

There are different definitions of C, one of which is just R^2 with a certain way of multiply stuff. In that definition, it's not exactly R, it's the set of guys that look like (x, 0), which is the same thing as R for all practical purposes. Any other definition of C will have a similar copy of R that is that same as R for all practical purposes.

 

[WWGD]

Each of C, R have different structure ; as sets, as fields, as metric spaces, etc. If you consider both as sets, then yes; R is a subset of C, it is the subset of the form {a+i.0: a is a Real number}. But there are relationships between the two other than this one.

 

[HallsofIvy]

Well, the addition and multiplication on C arace given by "(a+ bi)+ (c+ di)= (a+ c)+ (b+ d)i" and "(a+ bi)(c+ di)= (ac- bd)+ (ad+ bc)i". If we can think of the real number x as "x+ 0i". In that case (x+0i)+ (y+0i)= (x+ y)+ (0)i and (x+ 0i)(y+ 0i)= (xy- 0)+ (x0+ 0y)= xy+ 0i again. But since "(a, 0)" represents but is not the same as "a", rigorously, "The field of real numbers is isomorphic to a subfield of the field of complex numbers" rather than being a subset itself. Similarly as metric spaces- the metric space of real numbers is "homeomorphic" to a subspace of the complex numbers.

 

[FactChecker]

The short answer is yes. The issue of whether it is the same or just isomorphic to a subset of C is a distraction. Any two people talking about the reals are really talking about two abstract things that have the same properties and are isomorphic. This is true whether they are talking about subsets of C or not. So one person can think of R being identical to a subset of C just as easily as two people can think of their two abstract concepts of R being identical.

 

[haruspex]

Each of C, R have different structure ; as sets, as fields, as metric spaces, etc. If you consider both as sets, then yes; R is a subset of C, it is the subset of the form {a+i.0: a is a Real number}. But there are relationships between the two other than this one.

As Halls and FactChecker point out, it's a lot stronger than merely being a subset in the set-theoretic sense. (In fact, in that sense, it's a matter of choice whether you consider it a subset, or even which subset!)
The point is that all the structure of C, as inherited by the subset we think of as R, is an exact match for the structure of R. The dyadic etc. operations all still work and produce the 'right' answers. In short, it is an isomorphism, which is as close as you can get to saying two independently defined mathematical entities are the 'same'.

homeomorphic, as it happens, I've just been arguing on another thread against thinking of C as being R² with some multiplication rules. That's fine for a visualisation, but it can convey the impression that C is some sort of two dimensional vector space. This gets quite confusing when later you find C is a scalar field over which vector spaces can be defined.

 

[Therodre]

Hi,
I fail to see why that would be confusing. C is naturally endowed with a real 2-dimensional Vector space structure, ans also with a field structure (thus a complex 1-dimensional vector space structure).

Although i would not say that two isomorphic structure are as close as it gets to being the same, i'd say that what is closest is "uniquely isomorphic" (which is the case in the situation you're referring to) which can happen for any construction of C if you work with fields (that are R algebras) with one marked element.

 

[johnqwertyful]

As Halls and FactChecker point out, it's a lot stronger than merely being a subset in the set-theoretic sense. (In fact, in that sense, it's a matter of choice whether you consider it a subset, or even which subset!)
The point is that all the structure of C, as inherited by the subset we think of as R, is an exact match for the structure of R. The dyadic etc. operations all still work and produce the 'right' answers. In short, it is an isomorphism, which is as close as you can get to saying two independently defined mathematical entities are the 'same'.

homeomorphic, as it happens, I've just been arguing on another thread against thinking of C as being R² with some multiplication rules. That's fine for a visualisation, but it can convey the impression that C is some sort of two dimensional vector space. This gets quite confusing when later you find C is a scalar field over which vector spaces can be defined.

But C IS a 2 dimensional vector space over R. All field extensions are vector spaces over their base fields.

 

[haruspex]

But C IS a 2 dimensional vector space over R. All field extensions are vector spaces over their base fields.

No, it induces a 2D vector space, but it is not the same as a 2D vector space. Part of the definition of C is the multiplication function *:CxC→C.

 

[economicsnerd]

Whether it's formally a subset depends how you define $\mathbb R$, and how you define $\mathbb C$. There are a zillion different ways to build up these two objects from scratch, and your answer depends on which of those zillion ways we choose.

What is true under any standard definition is that there is a unique subfield $\hat{\mathbb{R}}$ of $\mathbb C$ which is isomorphic to $\mathbb R$. If we're being loose, that's enough to answer "yes" to your question.

 

[Terandol]

No, it induces a 2D vector space, but it is not the same as a 2D vector space. Part of the definition of C is the multiplication function *:CxC→C.

Why do you object to people saying $\mathbb{C}$ is a 2d real vector space but not to people saying it is a field or a metric space etc.? All people in this thread have said is that $\mathbb{C}$ is a 2d real vector space, not that it doesn't have additional structure. There are several strucutures $\mathbb{C}$ has which a general field doesn't. For example you can prove the only connected locally compact topological fields are $\mathbb{R}$ and $\mathbb{C}$ so a perfectly good definition of $\mathbb{C}$ is the unique (up to canonical isomorphism) connected locally compact topological field $\mathbb{F}$ such that $\mathbb{F}\setminus \{0\}$ is connected. So would you say $\mathbb{C}$ is not a field, it just induces a field since the topology with the above properties is a part of the definition of $\mathbb{C}$?

This is a highly nonstandard way to use language. Wouldn't this make every definition of the form: "A structure X is a structure Y together with the following extra things" that can be found in most textbooks (even those on formal logic) nonsensical since X is not Y it only induces Y. For example, I opened Lang's book on algebra to a random page and got the definition "A ring is a set A, together with two laws of composition..." Is Lang wrong when he says a ring is a set rather than something which induces a set if a forgetful functor is applied? If you do not object to this definition, what is wrong with defining $\mathbb{C}$ as a 2-dimensional real vector space together with a bilinear map satisfying some things plus a topology satisfying some things? Surely if this is a valid definition of $\mathbb{C}$ then one of the things $\mathbb{C}$ is, is a 2d real vector space which is all the post you are objecting to says.

 

[WWGD]

As Halls and FactChecker point out, it's a lot stronger than merely being a subset in the set-theoretic sense. (In fact, in that sense, it's a matter of choice whether you consider it a subset, or even which subset!)
The point is that all the structure of C, as inherited by the subset we think of as R, is an exact match for the structure of R. The dyadic etc. operations all still work and produce the 'right' answers. In short, it is an isomorphism, which is as close as you can get to saying two independently defined mathematical entities are the 'same'.

homeomorphic, as it happens, I've just been arguing on another thread against thinking of C as being R² with some multiplication rules. That's fine for a visualisation, but it can convey the impression that C is some sort of two dimensional vector space. This gets quite confusing when later you find C is a scalar field over which vector spaces can be defined.

I am neither saying nor implying that there aren't stronger relations other than that of being a subset of. I am saying C has many different types of structure, one being a set, and , if you isolate this aspect or this type of structure, then the Reals are a subset of C. Since the OP is somewhat ambiguous, I am making assumptions in answering. This does not mean nor imply that there aren't additional stronger relations between the two; I actually qualified my initial reply stating that there where additional relations between the two. As qualified, I still think what I said is correct .

And maybe if you want to define the restriction of C to R as being the same, you may want to appeal to some cathegory-theoretical or 'Mathematical-Logical' argument to make this claim more precise.

 

[homeomorphic]

I fail to see why that would be confusing. C is naturally endowed with a real 2-dimensional Vector space structure, ans also with a field structure (thus a complex 1-dimensional vector space structure).

Well, maybe I could see it being confusing, but it's a confusion you work through, and then once you sort it out, you realize it's not really confusing at all. I don't think you can side-step that confusing by not using the R^2 definition. I think it's something you have to work through, regardless.

 

[disregardthat]

Hi,
I fail to see why that would be confusing. C is naturally endowed with a real 2-dimensional Vector space structure, ans also with a field structure (thus a complex 1-dimensional vector space structure).

Although i would not say that two isomorphic structure are as close as it gets to being the same, i'd say that what is closest is "uniquely isomorphic" (which is the case in the situation you're referring to) which can happen for any construction of C if you work with fields (that are R algebras) with one marked element.

It's important to note that "Uniquely isomorphic" is really just isomorphic in disguise. When talking about an object classified up to unique isomorphism, we are referring to additional structure for which the object (together with this structure) is unique up to isomorphism. The unique part cones into play when talking about the associated universal property. However, all isomorphic object satisfy the same universal property.

For example, given a diagram functor in a category containing colimits, the colimit is unique up to unique isomorphism. What we mean here is that a colimit is a cone to the diagram (a cone is an object with an arrow to the diagram functor) which has the universal property that all other cones factors uniquely through it. So the colimit is just an initial object in the category of cones. But initial objects are only classified up to isomorphism. The object itself is not unique, but any arrow from it is unique.

 

[Therodre]

Well you're exemple looks confusing to me because as an exemple of object that is unique up to isomorphism you take an object that is unique up to unique isomorphism.
My point was that when objects are uniquely isomorphic then there's no harm in identifying them. Your example is a good illustration of that because initial objects are always unique up to unique isomorphism.
But some objects, such as the algebraic closure of a field for instance, are unique up to a non necessarily unique isomorphism, and it's sloppier to identify them.
I might have missed your point though (especially, when you say that all isomorphic objects will satisfy the same universal property as any of them does. You're right, but to satisfy a universal property you already need to be unique up to unique isomorphism)

 

[WWGD]

One point to consider is that using the term $\mathbb R$ may be ambiguous, in that there are many ways in which $\mathbb R$ may be made into a subset/subspace/etc. of $\mathbb C$ , with, AFAIK, none of these embeddings being somehow canonical. Sorry if that was the point that was being made to me, and I did not get it then. I think the OP needs to be pinned down more carefully to have a chance at a clear answer.

 

[FactChecker]

but to satisfy a universal property you already need to be unique up to unique isomorphism

I don't think this is correct. I bet we can come up with examples of things that are perfectly well defined but are not even uniquely isomorphic to themselves. Isn't the complex plane isomorphic to itself by the identity mapping and also by the mapping of each number to its conjugate? In fact, any nontrivial automorphism will probably be a good example. An isomorphism is enough to call things identical, unique or not.

 

[disregardthat]

I don't think this is correct. I bet we can come up with examples of things that are perfectly well defined but are not even uniquely isomorphic to themselves. Isn't the complex plane isomorphic to itself by the identity mapping and also by the mapping of each number to its conjugate? In fact, any nontrivial automorphism will probably be a good example. An isomorphism is enough to call things identical, unique or not.

Right. In the case for colimits, what's unique up to unique isomorphism is the cone (object with a set of morphisms), not the object in isolation. E.g. for any automorphism of the object, which there may be many of, only the identity will preserve the cone structure.

 

[Therodre]

Something can certainly be well defined and have many automorphism (and thus not be unique up to unique isomorphism) the field of complex number is typically such a thing. But such a thing will not satisfy a universal property in the corresponding category.
An object that satisfies a universal property is by definition always initial or final, and such objects are easily seen to be unique up to unique isomorphism.
Other objects are defined by choosing something at some point that is extraneous to the property we want it to have (either by specifically making a choice or "enriching" the category we work with).
The field C is well defined (in fact it has many possible definitions) it has many auto morphisms, however it also has a favored topology. For this topology only 2 automorphism are continuous that,s why when someone has two definitions (well defined of course) of C there are in general two isomorphisms between them, if we mark one element, that is a square root of -1, then it becomes unique up to unique isomorphism.
Of course you can always argue that C IS R2 with specific multiplication, and all other ways to define an algebraic closure of R define fields that are isomorphic to it.

 

[disregardthat]

It is not entirely correct that objects satisfying a universal property is unique up to unique isomorphism. The initial object (containing additional information) always belongs in a certain comma category, and while specifying an object in the category in question, it does not specify it up to unique isomorphisn.

 

[Therodre]

I had not seen disregardthat's answer.
I have to disagree again, you say that a colimit is only unique up to unique isomorphism (uui in short) as a (co-)cone. Nut a colimit is a co-cone. And certainly not only the object to which the arrow point, it is the object together with the arrows. If L is a colimit of a diagram in A, then forget(L) is not uui, that,s true but L has more structure than that it is an object to Co-Cone_D(A) (where D is the diagram indexing the cones).
To take a more down to Earth example. I'm sure that you would not claim that the tensor product A otimes B is just the module "naked", the bilinear map AtimesB to A otimes B is implicit in this picture.

 

[disregardthat]

Whenever the bilinear map is implicit you are in a different category. Namely in the category of pairs (P, AxB --> P) where P is a module and AxB --> P is bilinear. Here the tensor product (with its bilinear map) is initial and uui. But seen as a module, it is not. This was really my point.

 

[economicsnerd]

I have to wonder whether this discussion is the right one for clarifying the OP's question.

 

[micromass]

I have to wonder whether this discussion is the right one for clarifying the OP's question.

No, completely not. But the OP hasn't posted anything in this thread, so I assume he's gone. If he would post some kind of follow-up response, then I would make some effort in keeping the answers basic.

 

[micromass]

In the case that the OP is still following the thread for all intents and purposes, yes $\mathbb{R}$ is a subset of $\mathbb{C}$. So unless you're doing some advanced course like analysis or set theory, this is the right answer to your question. And even if you do the advanced course, the answer would still be morally right in my opinion. You can forget the rest of the answers in this thread which will likely be confusing and just take the answer that $\mathbb{R}\subseteq \mathbb{C}$ is true.

 

[Matterwave]

In the case that the OP is still following the thread for all intents and purposes, yes $\mathbb{R}$ is a subset of $\mathbb{C}$. So unless you're doing some advanced course like analysis or set theory, this is the right answer to your question. And even if you do the advanced course, the answer would still be morally right in my opinion. You can forget the rest of the answers in this thread which will likely be confusing and just take the answer that $\mathbb{R}\subseteq \mathbb{C}$ is true.

A question of notation here. Why not say $\mathbb{R}\subset\mathbb{C}$ instead of using the symbol $\subseteq$?

 

[micromass]

A question of notation here. Why not say $\mathbb{R}\subset\mathbb{C}$ instead of using the symbol $\subseteq$?

Oh yeah, you're right. I could as well have used $\subset$. It's just that I rarely ever use strict inclusion even if I know the inclusion is strict. Furthermore, many mathematicians use $A\subset B$ to denote an inclusion such that $A=B$ is also possible. I despise this notation with all my heart, so I try to use $\subset$ as rarely as possible. :tongue:

 

[disregardthat]

A question of notation here. Why not say $\mathbb{R}\subset\mathbb{C}$ instead of using the symbol $\subseteq$?

Some people interpret them the same, while others treat the former as proper subset. In the latter view, its sort of two statements, or a stronger statement. But the focus is only on whether R is contained in C, not that C has non-real members.

 

[jostpuur]

My answer is that $\mathbb{R}\subset\mathbb{C}$ is true in the same way as $\mathbb{Q}\subset\mathbb{R}$ is true. If you are using very rigor definitions, then $\mathbb{Q}\subset\mathbb{R}$ is not necessarily true, but usually people don't want to complain about it. IMO this answer is very good, because it really hits what is essential, and I know that this point of view is difficult enough to not be agreed upon by all. I have encountered people who insist that $\mathbb{Q}\subset\mathbb{R}$ would be absolutely true with no ambiguities, but at the same time $\mathbb{R}\subset\mathbb{C}$ would only be true in a sense.