Talk:Cardinality of the continuum

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A longer lead would help. Geometry guy 22:19, 16 June 2007 (UTC)

I was suprised to find that this page didn't exist so I thought I better start it. Admittedly, set theory is not my area of expertise so I hope others will improve it. I would like to see a nice clean proof of the oft stated fact that $c = 2^{\aleph_0}$ . I can't think of one at the moment (nor can I find one). -- Fropuff 06:42, 2005 Mar 6 (UTC)

Okay, I found a proof that I like and have included it in the article. If anyone knows of a better proof, feel free to change it. I am curious to know whether or not there exists an explicit bijection between the real numbers (or the unit interval) and power set of the natural numbers which isn't horribly convoluted. I've certainly never seen one (convoluted or otherwise). -- Fropuff 20:50, 2005 Mar 7 (UTC)

For a slightly less convoluted map than the one in the article, note that each real number in [0,1] is represented by a sequence of bits by the usual binary expansion. so 2^N ≥ c, and the other inequality as in the article. This is perhaps more concrete, and more similar to the second injection, so I like it. Maybe I'll swap it in. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

I did think about that one, but you'd have to add a line explaining that the cardinality of unit interval is equal to c; otherwise it works. It's also probably important to emphasize that the above map is only an injection and not a bijection. -- Fropuff 17:08, 31 January 2006 (UTC)

The proof of Schroeder-Berstein does give you an explicit bijection, if you follow it through. The note in the Cantor–Bernstein–Schroeder theorem article claiming that the proof is "not constructive" is a bit misleading; "not intuitionistically valid" would be more precise. --Trovatore 06:06, 3 October 2005 (UTC)

[edit] alef-omega

The article says that c can be taken equal to alef_{omega1}. But surely ω1 = ω, and we know that c cannot be alef_ω. This can't be right. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

Oh, I know. Surely it is meant the first uncountable ω_1, rather than the countable ω1. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

Oh, now that I edit it, I see that ω_1 is indeed indicated in the tex, it just doesn't display very clearly. So, nevermind. -lethe ^{talk +} 16:06, 31 January 2006 (UTC)

[edit] There are beth-one transcendental numbers?

I'd like to see a proof that there are $\mathfrak c$ transcendental numbers, because I never have before. OneWeirdDude (talk) 00:50, 7 January 2008 (UTC)

OK, here's a bare-bones sketch; holler if you need more details. There are only $\aleph_0$ algebraic reals. Suppose the cardinality of the set of all transcendental reals were $\kappa<2^{\aleph_0}$ . Then the cardinality of the set of all reals would be $\aleph_0+\kappa$ , which equals $\kappa\,\!$ , which is less than $2^{\aleph_0}$ . Contradiction.

If you want an argument that gives an actual example of a bijection between the transcendental reals and the reals, that's going to be a bit more involved, I'm afraid, but I'm confident it wouldn't be too hard. --Trovatore (talk) 01:02, 7 January 2008 (UTC)

The sum or difference of two algebraic real numbers is algebraic. Thus the sum or difference of an algebraic number and a transcendental number is transcendental. Similarly, the product of a nonzero algebraic number and a transcendental number is transcendental. e is known to be transcendental. For every n in the natural numbers and α in the algebraic numbers, map ((n+1)·e)+α to (n·e)+α. Map every other transcendental number to itself. This mapping is a bijection from the transcendental numbers to the reals. JRSpriggs (talk) 03:04, 8 January 2008 (UTC)

Nicely done! I had a much more brute-force solution in mind. --Trovatore (talk) 22:17, 8 January 2008 (UTC)

[edit] What is "Intuitive Argument?"

we are using, so for simplicity, let us consider a binary real number. Each position in its decimal expansion may hold either a 0 or a 1, so the number ${\mathfrak c}$ of all possible ways to fill those positions must be $2^{\aleph_0}$ . Therefore, ${\mathfrak c}=2^{\aleph_0}$ .

A prove by analogy: Each position in radix-3 number may hold either a 0 or a 1 or 2. The number ${\mathfrak c}$ of all possible ways to fill those positions must be $3^{\aleph_0}$ . Therefore, ${\mathfrak c}=3^{\aleph_0}$ . :) --Javalenok (talk) 14:08, 26 May 2008 (UTC)