20 Mar 08
Originally posted by SwlabrI think the answer is yes, based on this:
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
Originally posted by SwlabrI am not aware that anyone has proven that N_2 is the size of the real numbers.
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
Originally posted by preachingforjesusWith N_1 you mean Aleph One?
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
2 edits
Originally posted by TheMaster37I think this is a consequence from Godel's incompleteness theorems... But I have to do some research.
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
2 edits
Originally posted by adam warlockDrat-forgot about that.
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
Originally posted by GregMYup!
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 🙂