A simple explanation as to why $\mathbb R$ is not a subset of $\mathbb C$ from the point of view of set theory?

Question

I came across this remark in my complex analysis course. I haven't studied set theory in my courses thus I'm looking for a simplified explanation that does not involve too much formalism. I tried looking for some answers but the greatest barrier is the specific terminology

Answer 1

Without the context of that "remark" I can only guess at what it means. Here's my guess.

I suspect the book/instructor assumed you knew about the real numbers and constructed the complex numbers by defining the arithmetic on pairs $(x,y)$ of real numbers so that $(x,y)$ behaved as you would expect $x+iy$ to.

Now $x$ is a real number, but not a pair of real numbers, so the real numbers are not a subset of the complex numbers. That's weird and annoying, so we agree (formally of informally) to think about $x$ as the pair $(x,0)$. That makes (a copy of) the real numbers a subset of the complex numbers.

Answer 2

From the perspective of set theory, everything is a set. $\mathbb{R}$ is a set, containing numbers, which are really other sets, mostly containing other sets, etc. It's sets all the way down.

There are a few ways to define $\mathbb{R}$ and $\mathbb{C}$, but all of them are "equivalent" in the sense that they have identical structure. They may contain different sets, but it essentially comes down to the same thing with different "labels" on the numbers underneath. A different set might represent the same number.

One common way to define $\mathbb{R}$ is through Dedekind cuts. Without going into it, a number is a particular partition of the rational numbers (note that you need to define the rational numbers!). So, in this sense, the elements of $\mathbb{R}$ are sets of subsets of $\mathbb{Q}$.

In fact, none of the elements of $\mathbb{Q}$ actually lie in $\mathbb{R}$! From a set theory standpoint, $\mathbb{Q} \not\subseteq \mathbb{R}$. However, you can "redefine" $\mathbb{Q}$ to be elements of $\mathbb{R}$ that correspond to elements of $\mathbb{R}$, so that $\mathbb{Q} \subseteq \mathbb{R}$.

Similarly, $\mathbb{C}$ can be defined a few ways. One way is to simply make it $\mathbb{R} \times \mathbb{R}$, i.e. the plane, and define operations on it. In this case, elements of $\mathbb{C}$ are ordered pairs of reals. In that sense, again, none of the elements of $\mathbb{R}$ is in $\mathbb{C}$. But again, we could redefine $\mathbb{R}$ to be the elements of $\mathbb{C}$ of the form $(x, 0)$.

Basically, the exact set contents aren't important; the structure is. We often develop more complex structures from simpler ones, and in doing so, we don't typically get a superset.

Answer 3

There is a very common mistake that I see often. Set theory has no "built in mechanism" for interpreting what we can call "the rest of mathematics". We can prove that there is a way of doing that. But this is not different than proving that an algorithm halts, between there and its actual machine-level implementation via different languages, there is a big gap, and the two things are essentially not the same.

My point is that set theory has no "ordered pairs" or "natural numbers" or "real numbers", or anything else. It only has sets. And we have canonical ways to define ordered pairs, and canonical ways to build the natural numbers, and the real numbers, and the complex numbers.

But we also have other ways. The important thing from the set theoretic-foundationalist perspective is that the implementation you chose satisfies some basic properties which determine that it is a copy of the natural numbers/real numbers/etc.

We are free to define everything in the canonical way, reach $\Bbb C$ and then redefine $\Bbb R$ to mean the image of the canonical embedding of the "canonical construction of $\Bbb R$". In that scenario, $\Bbb R$ is in fact a subset of $\Bbb C$.

Of course, the canonical way of building $\Bbb C$ from $\Bbb R$ involves either endowing $\Bbb R^2$ with a multiplication structure, or looking at $\Bbb R[x]/(x^2+1)$. In either case, the result is that $\Bbb C$ is not a set which includes the real numbers themselves, but rather has a "canonical embedding of them".

But this does not mean that this is the only way to do so. Not even remotely.