# What is an advatage of defining $\mathbb{C}$ as a set containing $\mathbb{R}$?

It is a theorem that every field with least upper bound property and Archimedean property is isomorphic to each other.

So it seems not necessary to define $\mathbb{R}$ exactly and we simply denote any field with least upper bound property and Archimedean property as $\mathbb{R}$, in general.

Until now, i have defined $\mathbb{C}$ as a product of $\mathbb{R}$, that is, $\mathbb{R}\times\mathbb{R}$. (Very precisely in this case $\mathbb{R}$ is not a subset of $\mathbb{C}$)

However, i found this definition a bit uncomfortable when i'm doing arguments which should be done successively from real to complex. (i.e Abstract integral theory)

So i defined it newly, so that $\mathbb{R}\subset\mathbb{C}$.

I found this definition really natural than the first one since under this definition, $\mathbb{R}$ is a (topological) subspace of $\mathbb{C}$.

Well, i think this is advantageous, but i'm not sure.

Is there an advantage of using this definition? And is it Okay to use this definition?

• You can now say that $0\in\Bbb C$ and that $\pi\in\Bbb C$ and that $e^{\pi i x}$ is a function from the complex numbers to the complex numbers. The question by the way, can be extrapolated to ask what is the advantages of having $\Bbb Q\subseteq\Bbb R$. – Asaf Karagila Jul 23 '14 at 18:59
• @Mike: Yes, it is semantics. It's the same as asking "Is $\Bbb R$ a Dedekind-completion of the rational numbers, or a metric completion of the rational numbers?" both are very different sets (in the former a real number is a set/pair of sets of rational numbers, in the latter it is an equivalence class of sequences of rational numbers) but have the same final properties. – Asaf Karagila Jul 23 '14 at 19:10
• Least upper bound property and Archimedean property? Isn't the latter a consequence of the former? – bof Jul 23 '14 at 19:49
• @Asaf Thank you. (Sorry for deleting my comment: it was a mistake. For posterity, I'd asked "Since there's a unique field embedding $\mathbb R \hookrightarrow \mathbb C$, is this not just semantics?" Or something like that.) – user98602 Jul 23 '14 at 21:00

In ordinary everyday mathematics we do indeed say that $$\mathbb N\subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset \mathbb C$$ are honest-to-Plato, true, ordinary set inclusions, such that, for example, any natural number is literally a member of each of the other sets.

This is completely unproblematic, uncontroversial and not even deserving of mention as long as we're talking about everyday ordinary mathematics.

The problem you sketch only arises when we want to translate everyday mathematics into formal, axiomatic set theory. People who first learn of the possibility of doing this very easily get the impression that we want to (or ought to want to) do this all the time -- or even that the set-theoretic formalization is "what mathematics really is" and that everyday mathematics is just some kind of imperfect sloppy approximation to the purely set-theoretic Eternal Truth.

This latter extreme is not really tenable if you think about it (just consider that people were doing mathematics for millennia before axiomatic set theory was invented, and it is absurd to claim that Cantor, Zermelo, et al. were for the first time able to see what had really been going on). And even the milder view that we should be looking to the set-theoretic formalization all of the time is not really how mathematics works. The formalization is something we do once, to convince ourselves that what we're doing doesn't entail any contradictions that aren't present in set theory, such that set theory contains all the foundational uncertainty we'll ever need to worry about. After convincing ourselves of that, we'll promptly forget about its detail and continue doing everyday mathematics we always did, treating $$\mathbb N\subset \mathbb C$$ as a true inclusion and so forth.

Now, for the real question. Usually the formalization is done by saying that there are homomorphic injections $$\mathbb N\to\mathbb Z\to\mathbb Q\to\mathbb R\to \mathbb C$$ which preserve all the structure we're interested in, and then whenever we translate (well, imagine translating, because one doesn't actually do this, you know) an everyday formula into formal set theory, we're supposed to insert "invisible" applications of these homomorphisms and their inverses at appropriate places that will keep things making sense.

This works well enough that it's what most people who explain the formalization imagine doing.

However, there's also the option of making the inclusions be actual inclusions at the formal set theory level, which it sounds like you have rediscovered. For reference, it would go something like this for the step from $$\mathbb R$$ to $$\mathbb C$$:

First define $$\mathbb{\hat C}=\mathbb R\times \mathbb R$$ and make it into a field in the usual way. Then notice that there's an injective field homomorphism $$\phi: x\in\mathbb R\mapsto \langle x,0\rangle \in \mathbb{\hat C}$$, and now define $$\mathbb C = \mathbb R \,\cup\, \{\mathbb R\}\times(\mathbb{\hat C}\setminus \phi(\mathbb R))$$ where the $$\{\mathbb R\}$$ factor just serves to make sure the union is disjoint at the "untyped" set theory level. Then define $$\psi : \mathbb C \to \mathbb{\hat C}$$ by $$\psi(z) = \begin{cases} \hat z & \text{if }z = \langle\mathbb R,\hat z\rangle \\ \phi(z) & \text{if }z \in \mathbb R \end{cases}$$ then observe that $$\psi$$ is (obviously) a bijection, pull the field structure back along $$\psi$$ from $$\mathbb{\hat C}$$ to $$\mathbb C$$, and prove that the resulting field structure on $$\mathbb C$$ makes it into a true field extension of $$\mathbb R$$.

and we could apply exactly the same technique on each step on the way from $$\mathbb N$$ up to $$\mathbb R$$, such that we get true inclusions everywhere.

Your question is then, if I understand you correctly, whether this construction is "preferable" to the usual use of invisible injections everywhere.

Personally I like it better. It gives a set-theoretic formalization that is closer to how ordinary mathematics work, and I happen to care about such esthetic points.

It also seems to be useful to do things this way if we actually want to formalize mathematics in a computerized proof checking system. Then the gritty details of making the inclusions work can be isolated in the proof of the theorem "there exists something with the properties that $$\mathbb C$$ is usually taken to have", including the property of being a superset of $$\mathbb R$$ -- whereas the invisible-injection solution has to be supported either by special-casing the parsing of formulas all over ordinary mathematics, or by having a very complex generic solution to getting them inserted in the right places.

However, outside the computer-formalization exception, the actual mathematical answer is that it ultimately doesn't matter. Because no matter whether we're doing one thing or the other while arguing that ordinary mathematics can be reduced to formal set theory, what happens at the end of the day is still that we forget the details and continue doing ordinary mathematics as we've done it all the time anyway.

• Well written. +1 – Asaf Karagila Jul 23 '14 at 20:38
• It's very nice! Thank you – user156562 Jul 23 '14 at 20:55

For all intents and purposes, $\Bbb R$ was already in $\Bbb C$.

In a given category (like say, the category of topological $\Bbb R$ algebras) there is a notion of isomorphism. When two objects are isomorphic, they are usually considered indistinguishable, or at least you acknowledge they are copies of each other and share all the same properties.

Since $\Bbb R$ ("the set") is topological-algebra-homeomorphic to $\Bbb R\times \{0\}\subseteq \Bbb C$ (I'm assuming the left half is where you are defining the real part of $\Bbb C$), there's no sense in drawing a distinction between the two things.

In other words, looking at the two isomorphic things and saying "well they aren't the same because they aren't equal as sets" is being overly discerning.

That's why we normally consider a field $F$ to be a subring of any $F$ algebra. $F$ couldn't literally be a subset of all $F$-algebras because you can make $F$ algebras that are disjoint sets. But what happens is that we identify $F$ with all of its copies sitting inside each algebra.