# The set of real numbers is a subset of the set of complex numbers?

So, I was taught that $\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}$.

But since the set of complex numbers is by definition $$\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},$$ doesn't this mean $\mathbb{R}\subseteq\mathbb{C}$, since for each $x \in \mathbb{R}$ taking $z = x + 0i$ we have a complex number which equals $x$?

Yes, $\mathbb R \subset \mathbb C$, since any real number can be expressed as a complex number with $b=0$ (as you state).

Strictly speaking (from a set-theoretic view point), $\mathbb{R} \not \subset \mathbb{C}$. However, $\mathbb{C}$ comes with a canonical embedding of $\mathbb{R}$ and in this sense, you can treat $\mathbb{R}$ as a subset of $\mathbb{C}$.

On the same footing, $\mathbb{N} \not \subset \mathbb{Z} \not \subset \mathbb{Q} \not \subset \mathbb{R}$. However, there is an embedding of $\mathbb{N}$ in $\mathbb{Z}$, and similarly an embedding of $\mathbb{Z}$ in $\mathbb{Q}$ and an embedding of $\mathbb{Q}$ in $\mathbb{R}$.

You may want to look at this post for more details.

• You have $\not\subset$ if you construct them one after another. But already the fact that there are several constructions possible (e.g. Dedekind cuts or Cauchy sequences for $\mathbb R$) these ZFC models of $\mathbb R$ and the otger number sets are often not what we intuitively mean. We can as well consider a an algebraically closed field $\mathbb C$ of characteristic $0$ given and identify $\mathbb R,\mathbb Q,\mathbb Z,\mathbb N$ as actual subsets thereof. Commented May 12, 2013 at 22:41
• @HagenvonEitzen All the different constructions of $\mathbb{R}$ rely on the fact that we have already constructed $\mathbb{N}$ before (?).
– user17762
Commented May 12, 2013 at 22:42

So, I was taught that $$\mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}$$.

In general, we're often told that $$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$$, but, alas, none of these containments are true.

For instance, $$\mathbb{Z} := \mathbb{N}^2/R$$, where $$R$$ is the equivalence relation defined by $$R = \{ ((a,b),(c,d)) \in \mathbb{N}^2\times\mathbb{N}^2 \mid a+d = b+c\}$$ (see here for full details). With this definition, it is clear that $$\mathbb{N}$$ is not a subset $$\mathbb{Z}$$. However, one can identify every natural number $$n$$ with the equivalence class $$[(n,0)]$$.

As far as the original question, it is not the case that $$\mathbb{R} \subset \mathbb{C}$$. However, the mapping $$x\in \mathbb{R} \longmapsto x+0i \in \mathbb{C}$$ is a ring homomorphism (and, hence, an embedding) and an isometry (notice that $$\vert{x-y}\vert = \begin{Vmatrix} (x + 0i)-(y+0i)\end{Vmatrix}$$). Thus, the real axis is a subfield of $$\mathbb{C}$$ that is isomorphic to $$\mathbb{R}$$.