What is the Representation of \omega^\omega in a Visual Form?

[Dragonfall]

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

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

 

[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

 

[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.

 

[Dragonfall]

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

 

[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.

 

[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.

 

[CRGreathouse]

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.

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!

 

[Dragonfall]

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

 

[CRGreathouse]

I was confusing ordinal exponentiation with cardinal exponentiation.

I've always found that bizarre.

 

[Dragonfall]

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

 

[CRGreathouse]

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

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

 

[Dragonfall]

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

 

[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...

 

[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