# Embedding of $\mathbb{R}$ in $\mathbb{C}$

The natural inclusion of $$\mathbb{R}$$ in $$\mathbb{C}$$ is the mapping $$f: \mathbb{R} \to \mathbb{C}, \; x \mapsto x + 0i.$$ Given this, is it completely accurate to say that $$x \in \mathbb{C}$$? Or would we rather say that we can identity $$x$$ with an element $$x + 0i$$ that lives in $$\mathbb{C}$$? I assume that te former is true, since we do write that $$\mathbb{R} \subset \mathbb{C}$$, but since the complex numbers are by definition those numbers we can write in the form $$a + bi$$, I am not completely sure of why this is.

This mapping, in other words, seems less of sending $$x$$ to $$x + 0i$$, but rather asserting that $$x + 0i = x$$, so the identity element in $$\mathbb{C}$$ is not $$0 + 0i$$, but rather $$0$$.

Am I thinking of this correctly?

• It is purely a formal choice. Most often, people assume without comment that $\Bbb R\subset \Bbb C$, but if you were rigorously constructing the real and complex numbers, for example, you would want to be more careful. Mar 14, 2020 at 20:11

As you've constructed it, no, it would not be accurate to say $$x \in \mathbb{C}$$ since you explicitly stated that $$f: \mathbb{R} \to \mathbb{C}$$.
• My confusion is this, though: if $\mathbb{R} \subset \mathbb{C}$, doesn't $x \in \mathbb{R}$ imply $x \in \mathbb{C}$? Mar 14, 2020 at 21:03
• -1: Stating $f\colon \mathbb{R}\to \mathbb{C}$ does not imply that $\mathbb{R}$ is not a subset of $\mathbb{C}$. For example, I can write down a function $f\colon X\to Y$ with $X = \{1,2\}$ and $Y = \{1,2,3\}$ by $f(x) = x$. When I do this, I am in no way claiming that $1\notin Y$. Mar 14, 2020 at 21:07
• But the question wasn't "is $\mathbb{R}$ a subset of $\mathbb{C}$". The question was "is $x$ an element of $\mathbb{C}$". If we allow $x \in \mathbb{C}$, there's a degree of freedom that wouldn't be there if we decided $x \in \mathbb{R}$. Mar 14, 2020 at 21:11
• IMO, $1 + 0i$ is not a Real (even if we don't bother writing the $0i$). In other words, the $1$ from $\mathbb{R}$ isn't the same as the $1$ from $\mathbb{C}$ -- they just behave identically under certain assumptions (namely, the assumption that $\sqrt{-1}$ is defined). Mar 14, 2020 at 21:21
• Let me try to explain again. There are two reasonable conventions: $\mathbb{R}\subseteq \mathbb{C}$ or $\mathbb{R}\not\subseteq \mathbb{C}$. Under either convention, we could write down a function $f\colon \mathbb{R}\to \mathbb{C}$ by $f(x) = x+0i$, and this would make sense. Under the first convention, $x\in \mathbb{R}$ implies $x\in \mathbb{C}$. So the statement "as you've constructed it, no, it would not be accurate to say $x\in \mathbb{C}$ since you explicitly stated that $f\colon \mathbb{R}\to \mathbb{C}$" is just wrong. Apr 22, 2020 at 16:07