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Let us define $\mathbb{C}$ by $$\mathbb{C} = \mathbb{R}^2.$$ Real elements of $\mathbb{C}$ are tuples $(x, 0)$. But $x \neq (x, 0)$, so $\mathbb{R}$ cannot be a subset of $\mathbb{C}$.

I also read the following definition. Let $$\mathbb{C} = \{ a + bi \mid a, b \in \mathbb{R},\ i^2 = -1 \}.$$ But there is no $i \in \mathbb{R}$ such that $i^2 = -1$, so where does $i$ come from? If $i \in \mathbb{C}$ then we are using $\mathbb{C}$ in definition of $\mathbb{C}$.

NOTE: Apparently, there are similar questions on MSE but I think that lhf's answer is the best and most concise. Also, title "R is a subset of C?" is much easier to google than "Is a+0i in every way equal to just a?" I don't want readers to be redirected.


marked as duplicate by user223391, user91500, draks ..., Eric Wofsey, Claude Leibovici Dec 10 '15 at 8:22

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No matter how you define $\mathbb C$, the crucial point is that $\mathbb C$ contains a copy of $\mathbb R$, which we still denote by $\mathbb R$, such that $\mathbb C = \mathbb R + i\mathbb R$.

In mathematics, it matters less what things are than how they behave. If two things behave exactly the same, they're taken to be the same thing, even if the set-theoretic constructions of each do lead to different sets. The technical term is that we consider mathematical objects up to isomorphism, where isomorphism depends on the context.

This is especially strong when the objects have a categorical characterization that says they are essentially the only possible object with such characteristics. In the case at hand, $\mathbb R$ is the complete ordered field and $\mathbb C$ is the algebraic closure of $\mathbb R$ (or the only algebraic extension of $\mathbb R$ or the only quadratic extension of $\mathbb R$).

  • $\begingroup$ Certainly we have $\Bbb R + i \Bbb R = \Bbb C$. However (and honestly I'm not sure if there's a good answer here), is there any property of the real line $\Bbb R$ that distinguishes it from other lines through $0 \in \Bbb C$? $\endgroup$ – Omnomnomnom Nov 30 '15 at 17:59
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    $\begingroup$ @Omnomnomnom, yes, the other lines are not subfields of $\mathbb C$. $\endgroup$ – lhf Nov 30 '15 at 18:01
  • $\begingroup$ right, thanks ${}{}{}{}$ $\endgroup$ – Omnomnomnom Nov 30 '15 at 18:40

It is not wrong for you to worry, and the same problem occurs elsewhere:

  • We might define $\Bbb Z$ from $\Bbb N_0$ as a set of equivalence classes of $\Bbb N_0^2$ where $(a,b)\sim (c,d)\iff a+d=b+c$. But then $\Bbb N\not\subset \Bbb Z$.
  • We might define $\Bbb Q$ as a set of equivalence classes of $\Bbb Z\times \Bbb N$ where $(a,b)\sim (c,d)\iff ad=bc$. But then $\Bbb Z\not\subset \Bbb Q$.
  • We might define $\Bbb R$ via Dedekind cuts, which are infinie subsets of $\Bbb Q$; or as equvalence classes of rational Cauchy seequences modulo zero sequences. In both cases, $\Bbb Q\not\subset \Bbb R$.

But in each of these cases we have a canonical embedding of the smaller object into the larger object which completely respects the algebraic structure. In the above examples we have

  • $\Bbb N_0\to \Bbb Z$, $x\mapsto [(x,0)]$
  • $\Bbb Z\to\Bbb Q$, $x\mapsto [(x,1)]$ (where $1$ is the element of $\Bbb Z$ of that name, so it is $[(1,0)]$ according to the previous line; note that the $[\ ]$ denote equivalence classes with respect to totally different equivalence relations though)
  • $\Bbb Q\to\Bbb R$, $x\mapsto \{\,t\in\Bbb Q\mid t<y\,\}$, respectively $x\mapsto [(x,x,x,x,\ldots)]$

Of course we finde the same for $\Bbb R\to \Bbb C$, be it that we define $\Bbb C=\Bbb R^2$ with suitable operations, or $\Bbb C = \Bbb R[X]/(X^2+1)$.

The canonical nature of these embeddings allows us to loosely identify the smaller set with its image. In order to be very formally correct, we can stitch the sets together accordingly: Assume we have two disjoint sets $A,B$ and an injective map $\iota\colon A\to B$, then we can consider the set $\hat B:=(B\setminus \iota(A))\cup A$ instead of $B$ and obtain a set that allows the very same constructions as $B$ does while at the same time having $A$ as a subset. However, it is more cumbersome to try to define e.g. addition on such a stitched set $\hat B$ than to define addition straightforwardly on $B$ and using the canonical embedding to view $A$ as a subset of $B$.

During introduction it may be very important to make the distinction between $A$ and $\iota(A)$, but once the desired properties have been established it is best to forget about $\iota$, in fact to forget about the "mechanics" of constructing the larger set. It certainly won't help a physicist in solving his Schrödinger equation if he had to recall that the complex numbers he works with are ordered pairs of equivalence classes of sequences of equivalence classes of ordered pairs of equivalence classes of ordered pairs of finite ordinals (or was it ordered pairs of Dedekind cuts of sequences of equivalence classes of ordered pairs of an equivalence classes of ordered pairs of finite ordinals and a finite ordinal?).

Another justification usually comes as afterthought:

  • Given $\Bbb N$ let $\Bbb Z$ be any ring that contains $\Bbb N$ as sub-semiring and such that there is no ring $Z$ with $N\subsetneq Z\subsetneq \Bbb Z$. It turns out any two such rings are canonically isomorphic - and we need the ordered-pair-construction only to show the existence of such a ring.

  • Similarly, let $\Bbb Q$ be any minimal field containing $\Bbb Z$. It turns out any two such fields are canonically isomorphic.

  • Similarly, let $\Bbb R$ be any minimal complete field containing $\Bbb Q$. It turns out any two such fields are canonically isomorphic.
  • Similarly, let $\Bbb C$ be any minimal algebraically closed field containing $\Bbb R$. Actually, this time there is no canonical isomorphism of any two such fields: What we know as $i\in\Bbb C$ may map to either of two roots of the polynomial $X^2+1$. But once we fix one of these roots and call it $i$, we may consider $\Bbb C$ as $\Bbb R+\Bbb R i$.
  • $\begingroup$ +1 for ordered pairs of equivalence classes of sequences of equivalence classes of ordered pairs of equivalence classes of ordered pairs of finite ordinals (or was it ordered pairs of Dedekind cuts of sequences of equivalence classes of ordered pairs of an equivalence classes of ordered pairs of finite ordinals and a finite ordinal?) $\endgroup$ – mweiss Nov 30 '15 at 18:53

From the second definition you gave, the real numbers are just the complex numbers with $b=0$.

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    $\begingroup$ But we used $\mathbb{R}$ in definition of $\mathbb{C}$, so now we can't redefine $\mathbb{R}$. $\endgroup$ – edubrovskiy Nov 30 '15 at 17:54
  • $\begingroup$ I don't understand your reasoning. $\endgroup$ – Brandon Thomas Van Over Nov 30 '15 at 17:56
  • $\begingroup$ Why the heck do we need to "redefine" R? We define integers, then we define "even" using the definition of integer. What the heck is wrong with that? We don't need to "redefine" integer just because we used in in the definition of a subset or a supraset. $\endgroup$ – fleablood Nov 30 '15 at 17:59
  • $\begingroup$ @fleablood edubrovskiy is saying that if you define $\mathbb{C}$ as the set of $a + bi$ (which is really just going to be implemented as an ordered pair, anyway, if you do it formally) then you can't redefine the real numbers to be the set of ordered pairs where the second coordinate is zero. $\mathbb{R}$ can't be two different sets. This answer needs more detail. $\endgroup$ – 6005 Nov 30 '15 at 18:47
  • $\begingroup$ we don't redefine. The first definition of Reals still holds, we define C in terms of R, and then we observe that R is the subset of C where the second coordinate is zero. That is not a definition. It's an observed property--- a fundamental property that we can use axiomatically. $\endgroup$ – fleablood Nov 30 '15 at 19:36

$x \ne (x,0)$ but $f: R \times 0 = \{(x,0)\} \rightarrow R$ via $f((x,0)) = x$ is a isoomorphism so $R \cong R \times 0 \subset R^2$

And actually, in a sense, (x,0) is the same thing as x. It's fundamental that we can take two (or more) real numbers put them in an ordered pair and that doing so and creating a third thing, the ordered pair, we aren't actually changing are modifying the two numbers themselves. For example in the statement y = 2x + 3 the x'es and the y'es are real numbers yet we feel no conflict in saying (0,3) is a solution. How is the zero in the ordered pair supposed to be the same thing as 0, the real number, (0, ~) $\ne$ 0. and (~,3) $\ne$ 3?

Except they are. To the degree that the abstract concept of real numbers equaling 0 and 3 exist and there is a set of real numbers, R, is defined. Then abstract concept of cross referencing them and (0,3) $\in$ RxR where the "0" is 0 and the "2" is 2 and the R of the Rx{c} is R, is also legitimate. We can get down into the semantics and/philosophy and/or constructivism as to what embedding the Rs into RxR and view Rx{c} as the "same" as R and what is R after all, but in the end. Rx{c} is isomorphic to R in a fundamental and intuitive way that preserves all concepts, functions, and properties that any distinctions of how Rx{c} isn't R is as obscure as the definition of what R "really" is in itself.

(x,0) is equivalent to x in the viewing of R crossed with the single trivial point 0, and that is enough.


$i$ is an element of C but $i$ isn't C. There is nothing wrong with defining a set constructively from previously defined elements. That's how we define the integers and natural numbers after all.


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