How can $\mathbb{R} \subset \mathbb{C}$?

Question

My book defines $\mathbb{C}$ as $\{(x, y) : x,y \in \mathbb{R} \}$. So even when the imaginary part is zero, the elements of $\mathbb{C}$ are ordered pairs, while the reals are not.

I'm not sure if I should make a different question for this, but my book says that the reals are embedded in $\mathbb{C}$. Does this simply mean that the reals are a subset? Orthat we have an injective function from $\mathbb{R} \to \mathbb{C}$?

How did you introduce the rational numbers? Is $\mathbb{N} \subseteq \mathbb{Q}$? — user159517, Jan 28 '17 at 15:20
Not exactly a duplicate, but the answers to What is an advantage as defining $\mathbb C$ as a set containing $\mathbb R$? have useful discussion of these matters. — hmakholm left over Monica, Jan 28 '17 at 15:24

score 3 · Accepted Answer · answered Jan 28 '17 at 15:26

This is one of the quirks that comes up when you formally define the number systems. Usually, it happens none of the number systems $\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$ is literally a subset of the next, if you look at the formal constructions.

But each one embeds in an obvious way into the next one, and so there is no real harm in pretending that each is already a subset of the next. Most of the time, mathematicians will talk as if the number systems are already embedded in each other, except when we are specifically looking at the formal constructions of the number systems.

For example, the map that takes $x \in \mathbb{R}$ and sends it to $x + 0i$ in $\mathbb{C}$ is an embedding. In particular, it is a field isomorphism, so any field property of the reals will also hold for the set $\{x + 0i : x \in \mathbb{R}\}$.

score 2 · Answer 2 · answered Jan 28 '17 at 15:20

2

When people write $\mathbb{R}\subset \mathbb{C}$, they really mean that that there is a ring isomorphism between $\mathbb{R}$ and $\mathbb{R}^\prime = \{(x,0):x\in\mathbb{R}\}$, and that $\mathbb{R}^\prime \subset \mathbb{C}$.

answered Jan 28 '17 at 15:20

parsiad

25,738

I don't know any algebra, but I did a quick google search and I think I got the general idea. Isthis synonimous with embedding? – Ovi Jan 28 '17 at 15:24
@Ovi: that's correct – parsiad Jan 28 '17 at 16:23

score 1 · Answer 3 · answered Jan 28 '17 at 17:26

Adding to that a little bit to the other answers:

The topology of $\; \mathbb{C}$ is often identified with the topology of $\mathbb{R}^2$. We consider the isomorphism: $$ \pi:\mathbb{R}^2\to\mathbb{C}, (x,y)\to x+iy $$
where our norms of choice is often $||z||_\mathbb{C}^2=Re(z)^2+Im(z)^2$ and $||(x,y)||_\mathbb{R^2}^2=x^2+y^2$. And from that, you get that $\mathbb{R} \subset \mathbb{C}$, since $\mathbb{R}=\pi(\big\{(x,0)|x\in\mathbb{R} \big\}) \subset \mathbb{C}$.

By using this isomorphism, you can proof a lot of things in $\mathbb{C}$ while "being in $\mathbb{R}^2$". That why your textbook "defines" $\mathbb{C}$ that way. However, note that those are different spaces after all!

How can $\mathbb{R} \subset \mathbb{C}$?

3 Answers3

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