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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?

John P.
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    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. – pancini Mar 14 '20 at 20:11

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As you've constructed it, no, it would not be accurate to say $x \in \mathbb{C}$ since you explicitly stated that $f: \mathbb{R} \to \mathbb{C}$.

In response to your closing comments: mappings don't assert anything. They're just instructions for generating something based on what input was provided. (Although the person doing the math could make such an assertion and use that mapping as a persuasive argument.)

Jordan
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  • My confusion is this, though: if $\mathbb{R} \subset \mathbb{C}$, doesn't $x \in \mathbb{R}$ imply $x \in \mathbb{C}$? – John P. Mar 14 '20 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 $1\notin Y$. – Alex Kruckman Mar 14 '20 at 21:07
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    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 $x \in \mathbb{R}$. – Jordan Mar 14 '20 at 21:11
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    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). – Jordan Mar 14 '20 at 21:21
  • @AlexKruckman: I think he's saying "1" by itself is not in C. "1 + i0" is in C. I also see what you are trying to say too. The problem is that there is a matter of "opinion" in this debate. You can say "1" is in C" or that it's not because it doesn't have the imaginary component written down. If we just replaced R and C with more unusual objects, then in the category theory sense, you wouldn't say that the species x in A is the same as the species x in B. You can see under the user's name that they are a "New Contributor" so the downvote is quite harsh. – Nike Dattani Mar 14 '20 at 21:47
  • @user1271772 The point of view "The $1$ from $\mathbb{R}$ isn't the same as the $1$ from $\mathbb{C}$" is a totally reasonable one. My comment was just pointing out that "it would not be accurate to say $x\in \mathbb{C}$, since you explicitly stated that $f\colon \mathbb{R}\to \mathbb{C}$" is not a meaningful justification of this point of view. "If we allow $x\in \mathbb{C}$, there's a degree of freedom that wouldn't be there if we decided $x\in \mathbb{R}$" is also nonsense, as far as I can tell. – Alex Kruckman Mar 14 '20 at 22:04
  • @AlexKruckman: It makes sense to me: x in C means that x = 1 + i0. This has an "extra degree of freedom" that makes it impossible for it to be in R, because "1 + i0" is not in R! – Nike Dattani Mar 14 '20 at 22:06
  • @AlexKruckman I'm glad you agree with me! I would appreciate it if you would remove the downvote :). – Jordan Apr 22 '20 at 14:48
  • Sorry, my comment about your answer stands, and so does my downvote. – Alex Kruckman Apr 22 '20 at 15:48
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    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. – Alex Kruckman Apr 22 '20 at 16:07
  • I see, yes that makes sense. I think you have since edited your original comment, no? W/e, doesn't matter ^_^. – Jordan May 26 '20 at 01:16