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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 $x$?

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

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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.

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    $\begingroup$ 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. $\endgroup$
    – Hagen von Eitzen
    May 12, 2013 at 22:41
  • $\begingroup$ @HagenvonEitzen All the different constructions of $\mathbb{R}$ rely on the fact that we have already constructed $\mathbb{N}$ before (?). $\endgroup$
    – user17762
    May 12, 2013 at 22:42

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