Ordinal Exponentiation Comparison

[aa]

Which is bigger, a^b or b^a? (set theory)

Hi!

Thanks for letting me join your physics forums!

Will anyone help me with a set theory question I have? I've been racking my brains over this for the last two hours with no progress.

Which is greater using ordinal exponentation: \omega^{\omega_1} or \omega_1^{\omega}?

P.S. I know that \omega^{\omega_1} equals the order type of \underbrace{ \omega \times \omega \times \omega \times ... }_{\omega_1 \ many \ times}, and \omega_1^{\omega} equals the order type of \underbrace{ \omega_1 \times \omega_1 \times \omega_1 \times ... }_{\omega \ many \ times}, but I'm still stuck.

 

[dextercioby]

It depends,they can be equal even (for the a=b case and for 2 & 4,for example).So you can't formulate a general rule...

Daniel.

 

[aa]

Yes, but this is referring to ordinal exponentiation, a la set theory.

I really appreciate any help you all can give.

 

[Hurkyl]

What exactly is ω_1?

 

[aa]

Sorry, I see what you mean. This is the problem as given to me on the homework, but I have edited my post to use aleph notation. Is it a well-defined question now?

 

[Hurkyl]

Ack, alephs are usually used for cardinals aren't they? They're not order types! And cardinal exponentiation is different than ordinal exponentiation!

You could just say that ω_1 was the first ordinal with cardinality greater than that of ω.

 

[aa]

Hehe, alright, thanks. That's exactly what \omega_1 is supposed to be.

 

[Hurkyl]

Well, ω₁ is a limit ordinal, so what does that tell you about ω^ω₁?

 

[aa]

Well, ω₁ is a limit ordinal, so what does that tell you about ω^ω₁?

\omega^{\omega_1} = sup\{\omega^\alpha : \alpha < \omega_1\}. I did think of that, but don't see how it helps.

 

[Hurkyl]

What's the cardinality of &omega;^&alpha; if &alpha; < &omega;_1?

 

[aa]

It's \omega. So then \omega^{\omega_1} is an ordinal with cardinality equal to the union of \omega_1 many countable ordinals, which means \omega^{\omega_1} is an ordinal with cardinality \omega_1.

So we've shown that |\omega^{\omega_1}| = \omega_1, but it could still be true that \omega^{\omega_1} &gt; \omega_1.

 

[Hurkyl]

Better yet, it's a specific ordinal with cardinality &omega;_1.

Don't think of it in terms of unions, think of it in terms of the ordering...

 

[aa]

Oh schnap! You're right! \omega^{\omega_1} = \omega_1.

Boy, that's really surprising to me.

Thanks Hurkyl!

That IS what you had in mind, right? I'm not sure what the ordering has to do with it, but I do think that it = \omega_1. Now, anyway.

 

[Hurkyl]

Well, I thought of it in terms of the ordering: ω_1 is an upper bound of that set, so we must have \omega_1 \geq \omega^{\omega_1}.

 

[aa]

Hmm, let me think about it again for a sec...this is kinda tricky...

 

[Hurkyl]

You were right -- I didn't mean to suggest otherwise. Just offering how I arrived at that conclusion!

 

[aa]

Yes, thank you, and I am interested in your thought process for arriving at this conclusion too. It's just that after reading your post, I tried to think about it again, and suddenly wasn't sure of what I was thinking before.

Let me try to get this straight once and for all.

Your reasoning Hurkyl, is that for any countable ordinal \alpha, \omega^{\omega_1} &gt; \alpha because \omega^{\omega_1} &gt; \omega^{\alpha} &gt; \alpha. We do need to use the fact that \omega^{\alpha} &gt; \alpha for all countable ordinals \alpha, right? That seems true, although a proof isn't immediately obvious to me.

But in any case, the basic idea is that \omega^{\omega_1} &gt; all countable ordinals, right?

 

[Hurkyl]

I know \omega^{\omega_1} is bigger than any countable ordinal, because |\omega^{\omega_1}| = |\omega_1|. That's how I know that \omega_1 \leq \omega^{\omega_1}.

The other direction, \omega_1 \geq \omega^{\omega_1}, comes from the fact that \omega_1 is an upper bound of that set that defines \omega^{\omega_1}.

 

[aa]

I have finished writing my solution and am ready to turn it in. Whew.

Thank you Hurkyl. I really appreciate your helping me through this problem.

 

[davidseed]

going back to which is greater a^b or b^a. ( i can't use the special symbols)

then assume a>b (without loss of generality)
then
if b=1 then a^b =a b^a =1 so a^b is greater

there are some special cases for a,b, <=3 which I leave you to find

but for all other cases
b^a is greater e.g. 4^5 is greater than 5^4

I Haven't got a proof

this info comes from an Excel spreadsheet

but I think that the proof lies in putting
e=a-b

then
a^b = (b+e)^b = use binomial expansion
b^a = b^(b+e) = b^b * b^e

approach for proof

start with e=1
(certainly true here)


then recast problem as b+e as b(1+x) where x=e/b ; 0<x<1

David

 

[Hurkyl]

The problem isn't about natural numbers -- it's about ordinal numbers.