2
$\begingroup$

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?

$\endgroup$
1
  • 7
    $\begingroup$ 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. $\endgroup$
    – pancini
    Mar 14, 2020 at 20:11

1 Answer 1

0
$\begingroup$

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.)

$\endgroup$
11
  • $\begingroup$ My confusion is this, though: if $\mathbb{R} \subset \mathbb{C}$, doesn't $x \in \mathbb{R}$ imply $x \in \mathbb{C}$? $\endgroup$
    – John P.
    Mar 14, 2020 at 21:03
  • $\begingroup$ -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 $1\notin Y$. $\endgroup$ Mar 14, 2020 at 21:07
  • 1
    $\begingroup$ 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 $x \in \mathbb{R}$. $\endgroup$
    – Jordan
    Mar 14, 2020 at 21:11
  • 1
    $\begingroup$ 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). $\endgroup$
    – Jordan
    Mar 14, 2020 at 21:21
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 22, 2020 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .