# Are the reals genuinely a subset of the complex numbers? [duplicate]

In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $$z=(a,b)$$ with $$a,b \in \mathbb{R}$$. The imaginary unit $$i$$ is then just a shorthand for the ordered pair $$(0,1)$$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $$(a,0)$$ behaves in pretty much the same way as the real number $$a$$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $$(5,0)>(3,0)$$ in the way we can write $$5>3$$. With this in mind, I ask the following questions:

1. Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?
2. If the complex numbers are constructed in some other way, then is it meaningful to write $$\mathbb R \subset \mathbb C$$?
• See also math.stackexchange.com/a/3556451/442 ... Are the integers a subset of the rationals? Are the rationals a subset of the reals? Jan 2, 2021 at 20:20
• The Integers are defined as a set of ordered pairs of natural numbers, the Rationals are defined as a set of ordered pairs of integers. So if you can say Natural numbers are subset of Integers, you can say Real numbers are subset of Complex numbers Jan 2, 2021 at 20:23
• This feels like an abstract algebra question. For example, can you define the natural numbers without referring to the addition function or the total order on them? I'm not sure. What about the rationals, or the reals? Perhaps it depends on the context one is working in. I think for the natural numbers, you can just define them using the Peano axioms and have no additional structure. I'm not sure if you can do the same thing for the rational numbers or the reals. Jan 2, 2021 at 21:39
• @Joe to your point about the ill-definedness of $(4,0)>(3,0)$, one can consider the real numbers to be the subset $\{(a,0) : a \in \mathbb{R}\}$ equipped with the lexicographic ordering. Is this natural? No. Does it give us a bona-fide subset that makes up a metric space isometric to to $\mathbb{R}$? Yes. Jan 3, 2021 at 1:47
• Here’s a question about whether the natural numbers are “really” a subset of the real numbers. This strikes me as essentially the same question. Jan 3, 2021 at 1:55

Here is slightly abstract point of view: When we say $$A\subseteq B$$, we don't actually mean that $$A$$ is literally a subset of $$B$$. The symbol $$\subseteq$$ and its cousins $$\subset,\supset,\subsetneq,\supsetneq,\dots$$ are sensitive to context. If $$A$$ and $$B$$ are considered as sets, then the subset symbols mean what you were taught that they mean. But when $$A$$ and $$B$$ bear some additional structure, like being groups, vector spaces, metric spaces, fields, etc., then $$A\subseteq B$$ means:

"There is a natural, i.e., in some sensible way unique, injective homomorphism $$i:A\longrightarrow B$$, and from now on we mean $$i(A)$$ when we write $$A$$."

What a homomorphism is depends on the exact kind of structure we're looking at. If it's groups, we mean a group homomorphism. If it's vector spaces, we mean linear maps. If it's topological fields (fields equipped with a topology such that addition, multiplication and inversion are continuous), then we mean continuous field homomorphism.

Some authors try to get around this overloading of $$\subseteq$$ by specifying that $$\subseteq$$ literally means a subset and $$\leq$$ means what I wrote above. But if you ask me, in the big picture, it's not at all helpful to distinguish between the two cases. I can't think of a single instance where we actually care that, for instance, $$\mathbb R$$ is not literally a subset of $$\mathbb C$$ according to the standard construction of $$\mathbb C$$. We only care about how things behave, not how they look. And $$\mathbb C$$ contains a unique subset which, if equipped with the restricted field operations, behaves exactly like $$\mathbb R$$. And it's just cumbersome to always write "$$\mathbb C$$ contains a subfield isomorphic to $$\mathbb R$$" instead of just writing "$$\mathbb C$$ contains $$\mathbb R$$". So we do the latter.

Also, here's a better, less ambiguous way to talk about the subject: let $$F$$ be a field. A subfield of $$F$$ is a pair $$(E,i)$$, where $$E$$ is a field and $$i:E\to F$$ is an injective field homomorphism (field homomorphisms are automatically injective, but injectivity is needed for other structures). "Classical" subfields in the sense of subsets which are also fields are subfields in this sense if we take the natural inclusion mapping as $$i$$. And if $$i$$ is clear from context, we just say $$E$$ to be a subfield of $$F$$. Take this definition and then just say $$\mathbb R$$ is a subfield of $$\mathbb C$$, without using the subset symbol. It's what we care about, anyway: that one is a subfield of the other.

• I think that what you're saying is essentially correct, but it requires a bit of care when $A$ and $B$ can be thought of as carrying several structures. It can also be confusing in cases where there is an injective homomorphism, but it's not in some sense canonical. For instance, under your definition the statement $\mathbb{C} \subseteq \mathbb{R}$ is true if we're thinking of $\mathbb{C}$ and $\mathbb{R}$ just as groups, but that statement looks unnatural. Jan 3, 2021 at 5:47
• @ChrisEagle I think you get around that by also requiring that the homomorphism $i$ should be clear from context. In this case $i:a\mapsto (a,0)$ is the default so it's safe to write $\mathbb{R}\subseteq\mathbb{C}$. Although even this is not ideal, as sometimes you see people say $\mathbb{Z}_3\subseteq\mathbb{Z}_{15}$ without worrying about which group homomorphism is being implied.
– user632577
Jan 3, 2021 at 6:05
• No, this answer is wrong or at least very confusing. The inclusion symbol almost always means literal inclusion as sets. How would you define subgroups, subfields, subspaces otherwise? The symbol is sometimes abused to mean embedding, but IMO this is wrong and leads to all sort of confusion. Jan 3, 2021 at 9:00
• It is not enough that there is a subfield of $\mathbb{C}$ which is isomorphic to $\mathbb{R}$. We also need the isomorphism to be unique. Luckily that too is true. Jan 3, 2021 at 9:43
• @freakish: there are several definitions of sub<structure> (let's use groups for now). The most concrete is: Let $(G,\circ,e)$ be a group. A subgroup is a group of the form $(H,\circ\vert_{H\times H},e)$, where $H\subseteq G$. A more abstract definition: a subgroup of $G$ is a pair $(H,i)$, where $i:H\to G$ is an injective group homomorphism. The most abstract definition just drops the $H$, since that information is already contained in $i$, and says that a subgroup of $G$ is an injective homomorphism $i:H\to G$. This encompasses the first definition via the natural inclusion mapping. Jan 3, 2021 at 9:49

This is actually one of the problems with the idea of building maths purely on sets alone as foundations. If we take it strictly and formally, we get various statements which may or may not be true, subject to just how we have or have not constructed a particular object. For example, it gets even worse than what you are talking about: in a purely set-theoretic construction, natural numbers are sets, too, and thus we can ask whether, say,

$$1 \subseteq 3$$

and the answer to this is "it depends on your set-theoretic construction"!

For me, what I suggest is that this problem is very reminiscent of one often seen in computer programming: in computers, we have something similar going on in that everything we work with - pictures, sound, text, whatever - ultimately gets represented by the same "stuff": bits. And thus, if one does not have safeguards in place, one can try to interpret, say, the bits corresponding to text as a picture, or a picture as text, or conversely. Of course, what you get will be mostly scramble and nonsense, but you can do it, and the computer won't care.

So to deal with this, we need some way to encode that semantic information - that these two pieces of bits are semantically different - into the language in question.

And the way that is handled in computer programming is to use programming languages that require a data type, to discourage the programmer from arbitrarily mixing of different sets of bits that are meant to represent different things. Data typing attaches a semantic tag to each bit of data to say that it should represent a picture or text or a number, say, and then you cannot, in the same program, freely mix the two.

Likewise, this concept is not unheard of in maths - "type theory" explores a whole array of foundational systems and languages that use something very similar, and indeed both of these fields of application are closely related - but it's not the "standard consensus" foundation for maths.

But were we to use a typed foundation, I'd suggest the answer would best be thought of as a "no, but": the real numbers are not a subset of complex numbers, but we have the "type coercion" rule

$$x \mapsto (x, 0)$$

which allows us to "upgrade" a real number, should it be combined with a complex number in an expression, to a complex number. Such rules often feature in programming languages as I just mentioned, too. In general, they must be defined along with the types in question, but are typically based on whether or not "natural" correspondences of the kind you are perceiving here, exist.

• Does seem to be something with computer people, don't know why people pretend to be "poor" error prone computers. I think maybe this is best answer I've seen thus far. Feb 8, 2021 at 12:07
• I think the best math answer is very simply this. Firstly complex and real are independent of the issue in question. Only what matters is definition of subset and thus set itself, and Cartesian product of two sets. Feb 8, 2021 at 12:10

There are a lot of "inclusions" of a similar character, for example $$\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R$$. It depends on your interpretation of what these sets are whether you regard $$\subset$$ as a genuine inclusion or if you regard $$\subset$$ as a notation indicating that there is a canonical embedding of the object on left hand side as a subobject of the object on the right hand side.

Frequently one starts on the level of axiomatic set theory and then successively constructs $$\mathbb N$$, $$\mathbb Z$$, $$\mathbb Q$$, $$\mathbb R$$ etc. The usual constructions produce an "ascending sequence" of algebraic objects which are not related by genuine inclusions, but the respective construction yields a new object plus an embedding of the original object into the new one. Recall the construction of $$\mathbb Q$$ as a set of equivalence classes of pairs in $$\mathbb Z \times (\mathbb Z \setminus \{0\})$$.

However, usually there are various alternative constructions supplying specific models of the sets $$\mathbb N, \mathbb Z, \mathbb Q, \mathbb R$$ etc. But I think most people have a more abstract (or, if you want, more intuitive) understanding of these objects. We know what their purpose is, which operations can be performed and which rules are satisfied. The specific construction is then fairly irrelevant. I would call that an axiomatic point of view. For example, the reals can be constructed based on $$\mathbb Q$$ via Dedekind cuts, via nested intervals or via Cauchy sequences in $$\mathbb Q$$, and certainly there are even more approaches. But I doubt the anybody imagines a real number as a Dedekind cut or as an equivalence class of Cauchy sequences.

Therefore the concrete model of an object as $$\mathbb R$$ is fairly uninteresting, it can be replaced without problems by any other model having the same properties.

In your case we start with some model of $$\mathbb R$$ and construct $$\mathbb C$$ as the set of real pairs. Then $$\mathbb R' = \mathbb R \times \{0\}$$ is a subfield of $$\mathbb C$$ which is canonically isomorphic to the original $$\mathbb R$$ and we may now work with the new model $$\mathbb R'$$ which is a genuine subset of $$\mathbb C$$.

If you do not like this replacement, be aware that the standard construction of $$\mathbb C$$ is just one possible approach (but a very transparent one). It can be easily modified as follows. The simple idea is to add new (non-real) numbers to $$\mathbb R$$ instead of embedding $$\mathbb R$$ into $$\mathbb R^2$$ (which produces the isomorphic subfield $$\mathbb R' = \mathbb R \times \{0\} \subset \mathbb C$$ and causes the "philosophical discussion" whether, or in what sense, $$\mathbb R$$ is a genuine subset of $$\mathbb C$$).

Given $$\mathbb R$$, the sets $$\mathbb R$$ and $$\mathbb R \times \mathbb R^*$$ (where $$\mathbb R^* = \mathbb R \setminus \{0\}$$) are disjoint. Define $$\mathbb C = \mathbb R \cup (\mathbb R \times \mathbb R^*)$$ and $$\phi : \mathbb C \to \mathbb R \times \mathbb R, \phi(\zeta) = \begin{cases} (x,0) & \zeta = x \in \mathbb R \\ \zeta & \zeta \in \mathbb R \times \mathbb R^*\end{cases}$$

This is a bijection. The usual complex field structure on $$\mathbb R \times \mathbb R$$ is then transferred via $$\phi$$ to our modified $$\mathbb C$$. Doing so, we get a genuine inclusion $$\mathbb R \subset \mathbb C$$. But, to be honest, I can't see that this has an added value. Anyway, our model of $$\mathbb C$$ is a real vector space with basis $$\mathcal B = \{1 \in \mathbb R, i = (0,1) \in \mathbb R \times \mathbb R^* \}$$. Noting that $$x\cdot 1 = x$$ for $$x \in \mathbb R$$ and writing $$y \cdot i = i y$$ for $$y \in \mathbb R$$ we get $$\mathbb C = \{x\cdot 1 + y\cdot i = x + iy \mid x,y \in \mathbb R\} .$$ Here the complex number $$x + i0$$ is literally the same as the real number $$x$$.

Using similar methods we can of course construct a chain of genuine set inclusions $$\mathbb N \subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C .$$

1.- As already stated in your question, no: formally we can't say the reals are even a subset of the complex numbers. Yet there are many way in which we can embed $$\;\Bbb R\;$$ into $$\;\Bbb C\;$$ in such a way that the basic characteristics of these two fiels are mantained, and one of the most usual isomorphisms (=embeddings respecting the algebraic structure) is precisely $$\;\phi:\Bbb R\to\Bbb C\;,\;\;\phi(r):=r+0\cdot i\;$$ , or if you prefer the other very usual definition, $$\;\phi(r):=(r,0)\;$$. In this manner the real numbers become not only a subset but, as said, a subfield of the field $$\;\Bbb C\;$$ .

2.- Yes, we can define $$\;\Bbb C\;$$ as the algebraic closure of $$\;\Bbb R\;$$ , getting a fields extension of degree two, and in which in a rather canonical way, the basis field $$\;\Bbb R\;$$ is embedded in a the extension field $$\;\Bbb C\;$$ . In the sense of a theorem by Artin, this is the only possible algebraic extension of degree two of a real closed field.

• Your $\phi:\mathbb R \rightarrow \mathbb C$ is an "embedding respecting the algebraic structure", but that's not the same as an isomorphism. (Maybe a monomorphism.) Also, I wonder what are some of the "many other ways in which we can embed $\mathbb R$ into $\mathbb C$ in such a way that the basic characteristics of these fields are maintained"? Apr 1, 2021 at 5:13
• @TorstenSchoeneberg I used the term "isomorphism" in the sense explained there, though I guess it should have been a typo as I usually don't unse that term but rather monomorphism, which is almost sure what I meant back there. About how many such embeddings are there such that...etc.: that could also be a baseless claim that the hell knows why I wrote. There are infinite ways of embedding $\;\Bbb R\;$ into $\;\Bbb R^2\cong\Bbb C\;$ (for example, as real vector spaces), but such that the basic field char's are mantained is probably not that many... Apr 1, 2021 at 8:15

1. yes. The real numbers can be "said to be" a subset of the complex numbers. Even if it is not a genuine subset, we can "identify" $$\mathbb R$$ with a subset of $$\mathbb C$$; after we do that, we say $$\mathbb R \subset \mathbb C$$.

2. depends. (a) If $$\mathbb C$$ is constructed together with its topology, then there is a unique continuous automorphism $$\sigma$$ of order $$2$$ of $$\mathbb C$$ and its fixed set $$\{z \in \mathbb C : \sigma(z)=z\}$$ can be identified with the reals. (b) You could conceivably construct $$\mathbb C$$ in some other way. If you do not know what are the closed sets, or what are the Borel sets, or some structure in addition to the field structure, then what? There are many, many automorphisms of order two. The fixed set of any one of them is isomorphic to $$\mathbb R$$. But there is no way to single out one of these copies of $$\mathbb R$$ from the others.

In case (b) would we say we have actually constructed $$\mathbb C$$? Or merely that we have constructed some field isomorphic to $$\mathbb C$$? If it is isomorphic to $$\mathbb C$$ up to a non-unique isomorphism, can we say it "is" $$\mathbb C$$?

Contrariwise, a field can be isomorphic to $$\mathbb R$$ in at most one way, so if we construct any of these, then we do say it "is" $$\mathbb R$$.

It depends on how you define real numbers.

$$\mathbb{R}$$ can be defined by a set of axioms (a totally ordered field with the section separation element postulate).

In this setting, the construction you referred to is one of the many possible instances (technically called models) of "the real numbers", because it satisfies those axioms. All proposition you can construct about $$\mathbb{R}$$ that are proven true or false from those axioms will be the same for all models of $$\mathbb{R}$$, so distinguishing between those is not needed for such statements.

The subset of $$\mathbb{C}$$ you mentioned also satisfies the axioms, so it can also be called $$\mathbb{R}$$, making the inclusion statement true.

If, on the other hand, you define $$\mathbb{R}$$ to be one particular construction, all other isomorphic instances will be "just isomorphic". This approach has little use though, so it is usually not preferred.

• I can't see a reason why this answer was downvoted. Jan 3, 2021 at 11:45
• I think the reals can be defined as the power set of the integers? Jan 3, 2021 at 11:49
• @marshalcraft not in a straightforward manner. The construction I am familiar with is as equivalence classes of Cauchy sequences of rationals, but I'm sure there are others. I think that yields that the cardinality of the reals is the same as the power set of the rationals (and thus the integers), but there is a lot of work involved in making the definition work. Jan 3, 2021 at 12:10

After defining complex numbers you can realize that the subset $$X=\{(r,0)\ |\ r\in\mathbb{R}\}$$ with inherited addition and multiplication is isomorphic to $$\mathbb{R}$$. Inherited metric agrees as well. Hence we can replace our standard $$\mathbb{R}$$ with $$X$$ and now the inclusion property holds with other properties preserved as well.

In other words $$X$$ is a model of real numbers as well. And that's exactly what we do.

we can't write $$(5,0)>(3,0)$$ in the way we can write $$5>3$$.

Of course we can. What exactly stops you from doing that? This only depends on how (and on what) the "$$>$$" is defined. In fact for the $$X$$ I defined earlier this is a well defined total order, which agrees with the classical ordering of reals.

Note that reals can be defined by axioms. We often use this axiomatic approach for reals, not concrete model, because we usually care about properties, not how exactly things are constructed. After all reals can be constructed in many ways, e.g. Dedekind cuts or metric completion of $$\mathbb{Q}$$ or as the $$X$$ I defined earlier.