# Does this look right to you?

1. Sep 29, 2008

### Dragonfall

http://en.wikipedia.org/wiki/Image:Omega-exp-omega.svg [Broken]

It says it's a representation of $$\omega^\omega$$. But I really think the picture is countable. In fact, it looks like $$2\omega$$.

Last edited by a moderator: May 3, 2017
2. Sep 29, 2008

### Coin

I would really like to know under exactly what rule the image was generated before answering for sure. Is there anyone who speaks French and can give us a translation of the Description underneath? :P

3. Sep 29, 2008

### gel

Looks like $\omega^\omega$ to me, which is countable.

$\omega^\omega$ is order isomorphic to NxN, where you compare the first element first and then the second. Hopefully you know that NxN is countable.

4. Sep 29, 2008

### Dragonfall

No. $$\omega * \omega$$ is countable. $$\omega^\omega$$ is NOT. Think about it, $$2^{\aleph_0}=2^\omega\leq\omega^\omega$$.

5. Sep 29, 2008

### CRGreathouse

The French is pretty easy:

C++ program included in the XML source. The spiral represents all ordinals less than $\omega^\omega.$ The first turn of the spiral represents the finite ordinals, that is, 1, 2, 3, 4, etc. The second turn of the spiral represents the ordinals of the form
$$\omega\cdot m+n:\omega,\omega+1,\omega+2,\ldots,\omega\cdot2,\omega\cdot2+1,\omega\cdot2+2,\ldots,\omega\cdot3,\omega\cdot3+1,\ldots,\omega\cdot4,\ldots.$$
The third turn represents the ordinals of the form $\omega^2\cdot m+\omega\cdot n+p$ and the others are likewise; all the turns representing the powers of omega.​

Certainly, the ordinals below $\omega^\omega$ are countable. This construction makes that clear: a finite number of integers suffice to detail any such number.

6. Sep 29, 2008

### Dragonfall

The set of all the ordinals below $$\omega^\omega$$ is $$\omega^\omega$$, which is uncountable. So this image should be "uncountable". But it appears to be, since it's made of countable union of countable sets.

7. Sep 29, 2008

### CRGreathouse

I can biject all ordinals of the form $\omega^n\cdot a_n+\omega^{n-1}\cdot a_{n-1}+\cdots+\omega\cdot a_1+a_0$ with the natural numbers. Just consider the height $n+\sum a_i$ of an ordinal below $\omega^\omega$ and list the ordinals of height 0, the ordinals of height 1, the ordinals of height 2, and so on. There are only finitely many ordinals at each height, and each ordinal below $\omega^\omega$ has a finite height.

Alternately, by http://www.c2i.ntu.edu.sg/AI+CI/Humor/AI_Jokes/InvalidProofTechniques.html: Give one ordinal below $\omega^\omega$ not on my list!

Last edited by a moderator: Apr 23, 2017
8. Sep 29, 2008

### Dragonfall

My bad. I was confusing ordinal exponentiation with cardinal exponentiation.

9. Sep 30, 2008

### CRGreathouse

I've always found that bizarre.

10. Sep 30, 2008

### Dragonfall

The fact that people confuse the two? I just wasn't paying attention.

11. Sep 30, 2008

### CRGreathouse

No, just that the two are so different even though both are 'natural' extensions of the finite concept.

12. Sep 30, 2008

### Dragonfall

The confusion rises from the fact that $$\omega = \aleph_0$$, so you'd naturally expect that $$2^\omega = 2^{\aleph_0}$$.

13. Sep 30, 2008

### CRGreathouse

But the operators are overloaded, so the "^" in "2^omega" is different from the "^" in "2^{aleph_0}". Yes, I get that. I just find it curious that the two function so differently. Ordinals aren't uncountable until epsilon_0, right? That's a whole lot of exponentiation...

14. Sep 30, 2008

### Dragonfall

Yes, it is curious how the notions of "order" and "size" diverge so dramatically past finite numbers.

If I recall correctly you find cardinals more intuitive than ordinals. That's very odd! No offense:P