Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have some notes about curious facts about ordinal numbers, for example that their addition is not commutative, multiplication is not distributive from the right hand side and that the exponent rule doesn't always hold. Also that some things that are undefined in analysis like $0^0, \infty^0, 1^\infty$ are actually defined for ordinal numbers. I know that there's been some investigations devoted to pursuing ordinal arithmetic along the lines of classical results in number theory for example.

Do you happen to know other curious facts about ordinal numbers (compared to facts in analysis or other)?

share|improve this question
    
Nice list, but the notation is so small (at least in my browser!) –  amWhy Nov 21 '12 at 22:32
    
I don't think there are many fun facts on ordinals, or maybe I just don't know what you want to hear... –  Asaf Karagila Nov 21 '12 at 22:36
    
John Baez (math.ucr.edu/home/baez) has been writing a series of posts on Google+ (plus.google.com/117663015413546257905/posts) on countable ordinals, which you may find interesting. He is a very engaging expositor. The latest one is on the Hercules-Hydra game: plus.google.com/117663015413546257905/posts/HSS2PMVENe3 (and see also plus.google.com/117663015413546257905/posts/PjyerUy3AeH ) –  Andres Caicedo Nov 21 '12 at 22:38
2  
Well for example other fun facts I know of is that Fermat's Last Theorem and Goldbach's Hypothesis are known to be false in ordinal number theory (Sierpinski [1950]). –  Nicolas Nov 21 '12 at 22:48
1  
As a sanity check, by "$\infty^0$" and "$1^\infty$", do you really mean for $\infty$ as a variable whose values range over the infinite ordinals? As opposed, e.g. to suggesting there is an ordinal named $\infty$ or that it is the only infinite ordinal? –  Hurkyl Nov 21 '12 at 23:55
show 2 more comments

3 Answers 3

1) Every countable ordinal can be written in a unique, canonical way, called the Cantor Normal form. It is basically like writing the ordinal "in base omega."

2) There is a fact in number theory that has a natural proof using (necessarily) infinite ordinals. The question is does every "Goldstein sequence" eventually converge to 0. A Goldstein sequence is, basically, a process where you take a number, write it in base 2, replace all of the 2's with 3's, then subtract 1. Then do this for base 3 into 4, etc. The answer is yes they all converge, as you can see here, and the proof is rather shocking for a number theorist.

share|improve this answer
    
Welcome Mike! :-) –  Asaf Karagila Nov 22 '12 at 5:16
    
Why do you restrict yourself to countable ordinals? Every ordinal has a Cantor normal form. –  Chris Eagle Nov 22 '12 at 18:14
add comment

Every countable ordinal is order isomorphic to some closed subset of (the rationals in) the closed unit interval.

Not really related to ordinals, but it's might be interesting: $\mathbb Q$ is universal for all countable order types, i.e. any countable totally ordered set embeds into $(\mathbb Q,\leq)$.

share|improve this answer
add comment

Given any mapping $f : \mathbf{Ord} \to \mathbf{Ord}$ which is strictly increasing and continuous at limit ordinals (such a mapping is called a normal mapping), there are arbitrarily large ordinals $\alpha$ such that $f ( \alpha ) = \alpha$.

  • For example, the mapping $\alpha \mapsto \omega^\alpha$ is normal. The least fixed point of this function is commonly denoted $\epsilon_0$, and has great importance to the proof theory of Peano arithmetic.

    (Of course, as $\alpha \mapsto \omega^\alpha$ has arbitrarily large fixed points, we can define a new mapping $$\alpha \mapsto \epsilon_\alpha = \text{the } \alpha^{\text{th}} \text{ fixed point of the above mapping}.$$ This new mapping is also normal, and thus has arbitrarily large fixed points. And by a diagonalisation process at limit ordinals we can continue this indefinitely.)

  • The mapping $\alpha \mapsto \aleph_\alpha$ is also normal, and thus has arbitrarily large fixed points. That is, there are infinite cardinals $\kappa$ that have $\kappa$-many (infinite) cardinals below them. Even more, it is consistent with ZFC that $2^{\aleph_0} = | \mathbb{R} |$ is such a cardinal.

share|improve this answer
    
Haven't I told you about that last one? –  Asaf Karagila Nov 25 '12 at 20:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.