# Infinity (and beyond...)

Swlabr
Posers and Puzzles 20 Mar '08 14:19
1. 20 Mar '08 14:19
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"ðŸ˜‰.

By the continuum hypothesis, N_1=2^N_0.

Is n^N_0=N_1 for all n>1 (n a Natural number)?
2. PBE6
Bananarama
20 Mar '08 14:59
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"ðŸ˜‰.

By the continuum hypothesis, N_1=2^N_0.

Is n^N_0=N_1 for all n>1 (n a Natural number)?
I think the answer is yes, based on this:

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! ðŸ˜µ
3. 20 Mar '08 15:54
Originally posted by Swlabr
Assuming the Axiom of Choice:
...
By the continuum hypothesis, N_1=2^N_0.
...
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.
4. TheMaster37
Kupikupopo!
20 Mar '08 20:30
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.
With N_1 you mean Aleph One?

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...
Baby Gauss
20 Mar '08 21:502 edits
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...
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 hypothesis is neither provable nor disprovable from the axioms of Zermelo-Fraenkel set theory with the axiom of choice.
6. 20 Mar '08 23:212 edits
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.

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.
7. 22 Mar '08 00:01
Originally posted by Swlabr
Or another, more general, question: Does there exist a set of cardinality greater than that of the real numbers.
The power set of the reals, of course, but that's not a very interesting answer. Try the set of all curves.
8. TheMaster37
Kupikupopo!
25 Mar '08 10:02
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.
Yup!

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 ðŸ™‚
9. 26 Mar '08 06:161 edit
swlabr, could that have something to do with the rainbow having a beard? or a picture with a mustache?