Proof for similarity between R X R, and R

  • Thread starter Thread starter calvino
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

The discussion centers on proving the similarity between the Cartesian product of the reals, R x R, and the reals themselves, R. The first problem involves establishing a one-to-one and onto mapping from R x R to R, while the second problem seeks to demonstrate that (R x R, +) is isomorphic to (R, +). Participants highlight that standard bijections exist but may not serve as isomorphisms of groups, especially considering the complexities of infinite sets and their cardinalities. The Generalized Continuum Hypothesis is mentioned as a critical assumption for certain proofs regarding the sizes of infinite sets.

PREREQUISITES
  • Understanding of infinite set theory and cardinality
  • Familiarity with isomorphisms and group theory
  • Knowledge of bijections and mappings in mathematics
  • Concepts of the Generalized Continuum Hypothesis
NEXT STEPS
  • Research theorems related to the sizes of infinite sets and cardinality
  • Study the properties of isomorphisms in group theory
  • Explore standard bijections between R x R and R
  • Investigate the implications of the Generalized Continuum Hypothesis on infinite sets
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone interested in set theory and the properties of infinite sets.

calvino
Messages
108
Reaction score
0
Proof for similarity between R X R, and R...

Well, there are actually two similar questions that I need help on. The first is to prove that for any infinite set R, there is always a 1-1, and onto mapping from RxR to R.

The second is to show that (RXR, +) ~ (R, +). Here R is the reals, and ~ symbolizes that the two are isomorphic.

I know the two questions are similar, which is why I grouped them together. I'm just not sure where to start. Any help would be great, and I'll be sure to post my progress. Thanks
 
Physics news on Phys.org
Well, for the second one, I suppose the Pi function defined by PI(a,b)=a is good, right? What exactly is there to prove, if one can come up with an example (suitable for both cases)?
 
I phrased the first question wrong. I decided to make one thread with all my problems (since I have quite a bit that I'm stuck on)...sorry for any inconvenience.

closed...see "ALGEBRA PROBLEMS"
 
No, that is not an isomorphism. In particular, it is not injective, as you can easily check. Now there is something of a "standard" bijection from RxR to R, but I highly doubt it is an isomorphism of groups. Moreover, this bijection relies on decimal expansions, so it wouldn't be much good for your first problem which deals with arbitrary infinite sets.

For the second problem, my only guesses would be to do something abstract, i.e. I can't think of how to construct an actual isomorphism between the two. But maybe you can take some very large group, e.g. an infinite product of something, and find a surjective homomorphism from this group onto R, and another one onto RxR, but both with the same kernel, and then apply the first isomorphism theorem.

It seems like quite a tough problem. The two spaces are not homeomorphic as topological spaces, they are not isomorphic as rings, they are not isomorphic as vector spaces, and RxR isn't even a field. So RxR and R are different in so many ways, it's counterintuitive that they'd be the same as groups.

For the first problem, do you know any theorems regarding the sizes of infinite sets. Like if R is an infinite set, and S is another infinite set strictly larger than R, then is it true that |S| > 2|R|? If so, perhaps you can show |RxR| < 2|R|.
 
Last edited:
AKG said:
For the first problem, do you know any theorems regarding the sizes of infinite sets. Like if R is an infinite set, and S is another infinite set strictly larger than R, then is it true that |S| > 2|R|? If so, perhaps you can show |RxR| < 2|R|.
This only works if you assume the Generalized Continuum Hypothesis. :)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K