0

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.

Asaf Karagila
  • 393,674
Dov
  • 352
  • 3
    Careful. Dense has a precise meaning. You are using the term incorrectly. What do you mean by 'the density of complex numbers'? – Michael Albanese Jan 02 '14 at 05:45
  • you can define density as the least cardinal $\kappa$ such that there is a dense set of cardinality $\kappa$. In this sense, it's just $\omega$ for both. – user40276 Jan 02 '14 at 06:37
  • I fixed the question to hopefully make it more meaningful – Dov Jan 02 '14 at 12:13
  • 1
    Since the complex numbers are isomorphic to $\mathbb{R}^2$ have a look at this: http://mathworld.wolfram.com/Plane-FillingFunction.html. This is an example of a map that takes $\mathbb{R} \rightarrow \mathbb{R}^2$, showing that, in some sense, they are the same size. – postmortes Jan 02 '14 at 12:34
  • On a side note, the claim that the second is $\aleph_1$ is known as the continuum hypothesis. – ocg Jan 02 '14 at 12:53
  • Nit-pick historical remark regarding Cantor devised 1:1 mappings to prove that the set of integers was the same cardinality as positive integers, odd, etc. It's more correct to say that Cantor introduced the idea of using 1:1 mappings as a way of describing a certain relationship that a pair of sets may or may not have with each other. – Dave L. Renfro Jan 03 '14 at 15:54

2 Answers2

3

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.

Community
  • 1
Asaf Karagila
  • 393,674
  • It has been proved that it is not possible to prove or disprove that $\aleph_1=2^{\aleph_0}$? – Beni Bogosel Jan 03 '14 at 17:19
  • @Beni: Yes, Gödel proved the consistency of that in 1938, Cohen proved the other direction in 1963. – Asaf Karagila Jan 03 '14 at 17:44
  • That is good to know (I wonder why my university teachers never told that...). Is it hard for someone which only knows the basics of logic to comprehend the ideas of the proofs you mention? (just to know if I want to try or not...) Are there some good works which explain those to an accessible level? – Beni Bogosel Jan 03 '14 at 20:44
  • @BeniBogosel: You might want to read a beginner's guide to forcing by Chow. I guess it might be one of the more accessible documents on Cohen's proof, although only you can tell whether that is accessible enough. – MvG Jan 03 '14 at 21:46
  • @Beni, that depends on what you call basics. It probably requires understanding of axiomatic set theory. – Asaf Karagila Jan 04 '14 at 09:30
2

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.

MvG
  • 42,596