Cantor's infinities, and the cardinality of reals vs. complex

Question

Cantor devised 1:1 mappings to prove that the set of integers was the same cardinality as positive integers, odd, etc. And he proved that reals are infinitely more dense.

As I recall he called the first order of infinity $\aleph_0$ and the second $\aleph_1$.

What is the order of infinity for complex numbers compared to reals? is it $\aleph_2$? If reals are infinitely dense compared to ordinal numbers, would the same relationship be true of complex numbers compared to reals? Between any two reals, since they are already uncountable, there are an infinite number of reals, so are there even more complex numbers whose real part falls "between those two points"? It seems to me that however uncountable reals are, complex would be "more" but if it's already uncountable I can't see how to make an argument one way or the other.

Answer 1

The cardinality $\mathbb R$ has not been proven to be $\aleph_1$. The assumption of that fact is known as the continuum hypothesis. That hypothesis has been proven to be independent of the common ZFC set of axioms, so it can be neither proven nor disproven using that system. Wikipedia uses $\mathfrak c:=\lvert\mathbb R\rvert$ to denote the cardinality of the set of real numbers, and I believe this to be a common symbol.

On the other hand, it can be shown that $\lvert\mathbb C\rvert=\lvert\mathbb R^2\rvert=\lvert\mathbb R\rvert=\mathfrak c$. One way to see this is by considering the decimal representation of any complex number. You could take digits alternating from the real and the imaginary part and interleave them to form a real number. Sure, the sequence of digits would be infinite, but the thought experiment still works.

Answer 2

Cantor proved that $|\Bbb R|>|\Bbb N|$. He later proved that $|\Bbb R|=|\Bbb R^n|$, for any natural number $n>0$. Setting $n=2$ we have a natural bijection between $\Bbb R^2$ and $\Bbb C$, therefore this concludes that $|\Bbb R|$ and $|\Bbb C|$ are the same.

Cantor proved, at the same time, that $|\Bbb R|=|\mathcal P(\Bbb N)|$, where $\mathcal P$ denotes the power set operation. If we continue over the course of history, Cantor proved that $$|\Bbb R|=2^{\aleph_0}.$$

On the other hand, $\aleph_1$ is defined to be least cardinality which is larger than $\aleph_0$, and $\aleph_2$ is the least cardinality larger than $\aleph_1$. Cantor tried to prove that $2^{\aleph_0}=\aleph_1$, it was his conjecture for years before even writing the $\aleph$ symbols. But as it turns out, this statement - from the basic axioms of set theory - cannot be proved, nor disproved.

It is important to point out two prominent mistakes in your post:

1. The first is the arbitrary use of $\aleph$ number, which is sadly enough not uncommon. The common misconception that $\aleph_1$ is defined as the cardinality of the real numbers. This leads to even stranger claims, like $\aleph_2$ being the cardinality of the complex numbers.

   I am not scolding you, I am just pointing these mistakes which are grave and common.

2. Assuming the axiom of choice, of course, every set can be well-ordered. This means that there is some linear order $\prec$ on the set $\Bbb R$ such that $(\Bbb R,\prec)$ is well-ordered. This well-ordering has absolutely nothing to do with the natural order of the real numbers. Most people feel that there should be some compatibility between $\prec$ and $<$, and therefore become surprised when they are told that the axiom of choice implies that such well-order exists.

   Cardinality is what we have when we strip the sets of any structures they may have. So the "density" of the order of $<$ and the well-ordering of ordinals have little to do with the fact that $\Bbb R$ is equipotent with some ordinals.