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Math Help - Ordinal exponentiation

  1. #1
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    Ordinal exponentiation

    Hi,

    Suppose we take the usual definition of ordinal exponentiation (sorry for the notation!):

    a^0 = 1
    a^s(b) = a^b x a where s(b) is the successor of b
    a^b = Ua^c where c < b and b is a limit ordinal

    Then for a = 0, b = 1, and c = omega we have:

    0^1+c = 0^c = 1 because 0^c is U0^d for d < c and 0^0 = 1.
    0^1 = 0 and 0^c = 1 so 0^1 x 0^c = 0

    Thus for certain values of a, b, and c:

    ~(a^b+c = a^b x a^c)

    Is this right? (Again an exercise says it should hold for all a, b, and c)

    Any help would be great.

    Regards
    Sam
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  2. #2
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    Nice catch. Wikipedia and PlanetMath give definitions with this clause: "if δ is limit, then α^δ is the limit of the α^β for all β < δ." However, MathWorld separates the case of the zero base: "If \beta is a limit ordinal, then if \alpha=0, \alpha^\beta=0. If \alpha\ne0 then, \alpha^\beta is the least ordinal greater than any ordinal in the set \{\alpha^\gamma:\gamma<\beta\}."

    The first Wikipedia definition in terms of ordered sets says the following.
    In general, any well ordered set B can be raised to the power of another well ordered set E, resulting in another well ordered set, the power B^E. Each element of B^E is a function from E to B such that only a finite number of elements of the domain E map to an element larger than the least element of the range B (essentially, we consider the functions with finite support).
    According to this, 0^\omega is the set of functions from \omega to \emptyset, i.e., \emptyset.
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  3. #3
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    I'd rather eschew definitions from unreliable Internet sites.

    The definition given in the first post is one that I am familiar with from standard textbooks in set theory.

    [EDIT: On checking, I find that I did not correctly remember those textbook definitions. My post 11 should clear up whatever confusion I caused.]
    Last edited by MoeBlee; September 24th 2010 at 07:36 AM.
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  4. #4
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    Well, doesn't the OP show that this definition is incorrect?

    The MathWorld cites two books.
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  5. #5
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    Quote Originally Posted by sroberts View Post
    Then for a = 0, b = 1, and c = omega we have:

    0^1+c = 0^c = 1 because 0^c is U0^d for d < c and 0^0 = 1.
    I take it you mean 0^(1+c). In that case your calculation is incorrect:

    0^(1+c) = 0^c = U_b<c 0^b = U{x | Eb<c x=0^b} = U{0} = 0.

    [EDIT: I'm was mistaken there. It's not the case that {x | Eb<c x=0^b} = {0}.]
    Last edited by MoeBlee; September 23rd 2010 at 12:57 PM.
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  6. #6
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    No. The poster's calculation is incorrect.

    [EDIT: I'm mistaken.]
    Last edited by MoeBlee; September 23rd 2010 at 12:58 PM.
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  7. #7
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    But 0^0 = 1 both from the definition by recursion and because there exists a single, namely, empty, function from 0 to 0. Therefore, \bigcup_{b<\omega} 0^b\supseteq 1.
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  8. #8
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    That was my thought emakarov. For every limit ordinal a, 0^a will be 1 because 0^0 is 1; and for every successor ordinal it will be 0.

    ...

    sam
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  9. #9
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    Right, I made a mistake.

    U_b<c 0^b = U{x | Eb<c x=0^b}.

    And both 0 and 1 satisfy x=0^b for some b<c.

    So U_b<c 0^b = U{0 1} = 1.

    So the poster is correct that 0^(1+c) = 1

    And the poster's original definition is the one that I recall from many the textbooks I've seen.

    But, as I recall, for all a, b, c, we have a^(b+c) = (a^b)x(a^c) .

    So something is wrong here. Something is being overlooked here, something not being stated or calculated correctly.

    What textbook is it, by the way? (Not that it should matter, though, as far as I can tell, since the puzzling situation here involves a standard definition and a well known distributivity theorem.)
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  10. #10
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    Hi,

    It's Just and Weese's discovering modern set theory again.

    Thanks for the help from both of you.

    Cheers
    Sam
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  11. #11
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    Thanks. I'll look at it.

    I checked some other textbooks last night. The definition you give is also given by Jech, but I coudn't find him also asserting a*(b+c) = a^b * a^c.

    On the other hand, the following textbook definitions are equivalent, and are different from the one in your original post (so I had misrembered the following definitions) [note: as formulated here, 0 is not a limit ordinal]:

    ENDERTON:

    If a>0 then:
    a^0 = 1
    a^b+ = a^b * a
    if b is a limit then a^b = U{a^c | c<b}

    also:
    0^0 = 1
    if ~b=0 then 0^b = 0
    ______

    LEVY:

    a^0 = 1
    a^b+ = a^b * a
    if b is a limit then a^b = U{a^c | 0<c<b}
    ______

    SUPPES:

    a^0 = 1
    a^b+ = a^b * a
    if b is a limit and a>0, then a^b = U{a^c | 0<c<b}
    if b is a limit then 0^b = 0
    ______

    Those are all equivalent and prove a*(b+c) = a^b * a^c.

    Jech's definition is different, but he does not assert a*(b+c) = a^b * a^c.

    So, given that you accurately restated the Just & Weese definition, it does seem that the exercise is incorrect.
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  12. #12
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    That makes a lot more sense.

    Thanks again,
    Sam
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