- 20 Mar '08 14:59

I think the answer is yes, based on this:*Originally posted by Swlabr***Assuming the Axiom of Choice:**

Let N_0 denote the size of the Natural numbers ("countably infinite), and let N_1 denote the size of the Real numbers ("uncountably infinite".

By the continuum hypothesis, N_1=2^N_0.

Is n^N_0=N_1 for all n>1 (n a Natural number)?

http://en.wikipedia.org/wiki/Continuum_hypothesis#Implications_of_GCH_for_cardinal_exponentiation

Of course, I don't have any formal instruction in set theory so I don't actually know what it means, but it fits the first pattern! - 20 Mar '08 15:54

I am not aware that anyone has proven that N_2 is the size of the real numbers.*Originally posted by Swlabr***Assuming the Axiom of Choice:**

...

By the continuum hypothesis, N_1=2^N_0.

...

n^N_1=N_2 : n= real | n>1*number of reals*N_R>N_1

these I'm familliar with; if anyone has proven N_2=N_R please give reference. - 20 Mar '08 20:30

With N_1 you mean Aleph One?*Originally posted by preachingforjesus***I am not aware that anyone has proven that N_2 is the size of the real numbers.**

n^N_1=N_2 : n= real | n>1*number of reals*N_R>N_1

these I'm familliar with; if anyone has proven N_2=N_R please give reference.

Then, yes, n^N_0 = N_1

I used to be able to prove it too (had good fun with Axiomatic Set Theory) , but the knowledge has been collecting dust for some time now :p

The Continuum Hypothosis claims that N_1 would be equivalent to the reals, so if someone proved that N_2 is equivalent to the reals that person would have a proof against CH.

I'm not sure, but I seem to recall a proof, proving that CH can't be proven or disproven in Set Theory... - 20 Mar '08 21:50 / 2 edits

I think this is a consequence from Godel's incompleteness theorems... But I have to do some research.*Originally posted by TheMaster37***I'm not sure, but I seem to recall a proof, proving that CH can't be proven or disproven in Set Theory...**

Edit: This is what I got after a quick search :"He also showed that the continuum hypothesis cannot be disproved from the accepted axioms of set theory, if those axioms are consistent"

Edit2: Later work by Paul Cohen established that the continuum hypothesis is neither provable nor disprovable from the axioms of Zermelo-Fraenkel set theory with the axiom of choice. - 20 Mar '08 23:21 / 2 edits

Drat-forgot about that.*Originally posted by adam warlock***I think this is a consequence from Godel's incompleteness theorems... But I have to do some research.**

Edit: This is what I got after a quick search :"He also showed that the continuum hypothesis cannot be disproved from the accepted axioms of set theory, if those axioms are consistent"

Edit2: Later work by Paul Cohen established that the continuum ...[text shortened]... able nor disprovable from the axioms of Zermelo-Fraenkel set theory with the axiom of choice.

How about asking,

We know N_0 < 2^N_0 (as CH asks if there exists and S s.t. N_0 < |S| < 2^N_0). Is n^N_0 > 2^N_0? (For all n>2, natural)

Or another, more general, question: Does there exist a set of cardinality greater than that of the real numbers. - 25 Mar '08 10:02

Yup!*Originally posted by GregM***The power set of the reals, of course, but that's not a very interesting answer. Try the set of all curves.**

I do have to mention that my set-theory is based on Zermelo-Fraenkel. It was a difficult subject at first, but after a while I got the knack for it wich made the subject infinitely more fun