Why $\mathbb{R}$ is a subset of $\mathbb{C}$? [duplicate]

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 $\mathbb R$).

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 $\Bbb R+\Bbb R i$.