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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 $x$?
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.
