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Which is bigger, a^b or b^a?

  1. Apr 13, 2005 #1

    aa

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    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: [tex]\omega^{\omega_1}[/tex] or [tex]\omega_1^{\omega}[/tex]?

    P.S. I know that [tex]\omega^{\omega_1}[/tex] equals the order type of [tex]\underbrace{ \omega \times \omega \times \omega \times ... }_{\omega_1 \ many \ times}[/tex], and [tex]\omega_1^{\omega}[/tex] equals the order type of [tex]\underbrace{ \omega_1 \times \omega_1 \times \omega_1 \times ... }_{\omega \ many \ times}[/tex], but I'm still stuck.
     
    Last edited: Apr 14, 2005
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  3. Apr 14, 2005 #2

    dextercioby

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    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.
     
  4. Apr 14, 2005 #3

    aa

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

    I really appreciate any help you all can give.
     
    Last edited: Apr 14, 2005
  5. Apr 14, 2005 #4

    Hurkyl

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    What exactly is ω_1?
     
  6. Apr 14, 2005 #5

    aa

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    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?
     
  7. Apr 14, 2005 #6

    Hurkyl

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

    You could just say that ω_1 was the first ordinal with cardinality greater than that of ω. :tongue2:
     
  8. Apr 14, 2005 #7

    aa

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    Hehe, alright, thanks. That's exactly what [tex]\omega_1[/tex] is supposed to be.
     
  9. Apr 14, 2005 #8

    Hurkyl

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    Well, ω1 is a limit ordinal, so what does that tell you about ω^ω1?
     
  10. Apr 14, 2005 #9

    aa

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    [tex]\omega^{\omega_1} = sup\{\omega^\alpha : \alpha < \omega_1\}[/tex]. I did think of that, but don't see how it helps.
     
  11. Apr 14, 2005 #10

    Hurkyl

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    What's the cardinality of &omega;^&alpha; if &alpha; < &omega;_1?
     
  12. Apr 14, 2005 #11

    aa

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

    So we've shown that [tex]|\omega^{\omega_1}|[/tex] = [tex]\omega_1[/tex], but it could still be true that [tex]\omega^{\omega_1} > \omega_1[/tex].
     
    Last edited: Apr 14, 2005
  13. Apr 14, 2005 #12

    Hurkyl

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    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...
     
  14. Apr 14, 2005 #13

    aa

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    Oh schnap! You're right! [tex]\omega^{\omega_1} = \omega_1[/tex].

    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 = [tex]\omega_1[/tex]. Now, anyway.
     
  15. Apr 14, 2005 #14

    Hurkyl

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    Well, I thought of it in terms of the ordering: ω_1 is an upper bound of that set, so we must have [itex]\omega_1 \geq \omega^{\omega_1}[/itex].
     
  16. Apr 14, 2005 #15

    aa

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    Hmm, let me think about it again for a sec...this is kinda tricky...
     
  17. Apr 14, 2005 #16

    Hurkyl

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    You were right -- I didn't mean to suggest otherwise. Just offering how I arrived at that conclusion!
     
  18. Apr 14, 2005 #17

    aa

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    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 [tex]\alpha[/tex], [tex]\omega^{\omega_1} > \alpha[/tex] because [tex]\omega^{\omega_1} > \omega^{\alpha} > \alpha[/tex]. We do need to use the fact that [tex]\omega^{\alpha} > \alpha[/tex] for all countable ordinals [tex]\alpha[/tex], right? That seems true, although a proof isn't immediately obvious to me.

    But in any case, the basic idea is that [tex]\omega^{\omega_1} > [/tex] all countable ordinals, right?
     
  19. Apr 14, 2005 #18

    Hurkyl

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

    The other direction, [itex]\omega_1 \geq \omega^{\omega_1}[/itex], comes from the fact that [itex]\omega_1[/itex] is an upper bound of that set that defines [itex]\omega^{\omega_1}[/itex].
     
  20. Apr 15, 2005 #19

    aa

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    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.
     
  21. Apr 21, 2005 #20
    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
     
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