Is 0^omega = 0 or 1?
Gerald
A question about ordinal exponentiation. In my opinion 0^a should be 0
for every nonzero ordinal number a. But the recursive definition of
ordinal exponentiation in all books and texts i found is stated as:
a^0 = 1
a^(b+1) = a^b * a
a^b = sup {a^c : c < b}, if b is limit (1)
Now this yields
0^0 = 1
0^n = 0 for 0 < n < omega
and thus by (1)
0^omega = sup {1,0} = 1.
This would mean that the function 0^a is not continous at a = omega.
It also contradicts the natural combinatorial definition of a^b:
Take the set A of all functions from b into a, such that f(x) is 0
outside a finite subset of b. Order A lexicographically as
f < g iff there exists an ordinal y < b with f(y) < g(y) but f(x) =
g(x) for y < x < b.
This wellorders A an its order type is supposed to be a^b. But here
clearly 0^omega = 0 as there is no function from omega into 0.
I think, that (1) should be corrected to
a^b = sup {a^c : 0 < c < b}, if b is limit
and that (1) is a typo. But (1) is present in all books about set
theory ive checked:
Jech "Set Theory"
Kunen "Set Theory: An Introduction to Independence Proofs"
Holz, Steffens, Weitz "Introduction to Cardinal Arithmetic"
David C. Ullrich
>Hi
>
>A question about ordinal exponentiation. In my opinion 0^a should be 0
>for every nonzero ordinal number a. But the recursive definition of
>ordinal exponentiation in all books and texts i found is stated as:
>
>a^0 = 1
>a^(b+1) = a^b * a
>a^b = sup {a^c : c < b}, if b is limit (1)
>
>Now this yields
>
>0^0 = 1
>0^n = 0 for 0 < n < omega
>
>and thus by (1)
>
>0^omega = sup {1,0} = 1.
Huh - very interesting.
I don't know what the official take on this would be, but my
feeling as a sort of educated layman in set theory is that
people would say that this example shows that the traditional
definition as above is not what people really mean. When
people say (1) they're overlooking this one case in which
the exponential is not monotone; they really mean for (1)
to read
a^b = lim_{c -> b from below} a^c , if b is limit (1')
but they write (1) instead because they overlooked this
one case where the two are not equivalent.
>This would mean that the function 0^a is not continous at a = omega.
>It also contradicts the natural combinatorial definition of a^b:
>
>Take the set A of all functions from b into a, such that f(x) is 0
>outside a finite subset of b. Order A lexicographically as
>
>f < g iff there exists an ordinal y < b with f(y) < g(y) but f(x) =
>g(x) for y < x < b.
>
>This wellorders A an its order type is supposed to be a^b. But here
>clearly 0^omega = 0 as there is no function from omega into 0.
>
>I think, that (1) should be corrected to
>
>a^b = sup {a^c : 0 < c < b}, if b is limit
I'd tend to agree if I felt entitled to have an opinion; that's
a simpler way to state the same "correction" I suggested above.
(Although I think the way I put it is better because it says
more directly what we "really mean" here...)
>and that (1) is a typo. But (1) is present in all books about set
>theory ive checked:
>
>Jech "Set Theory"
>Kunen "Set Theory: An Introduction to Independence Proofs"
>Holz, Steffens, Weitz "Introduction to Cardinal Arithmetic"
************************
David C. Ullrich
William Elliot
> On 2 Oct 2004 03:50:40 -0700, gerald...@gmx.de (Gerald) wrote:
> >
> >A question about ordinal exponentiation. In my opinion 0^a should be 0
> >for every nonzero ordinal number a. But the recursive definition of
> >ordinal exponentiation in all books and texts i found is stated as:
> >
> >a^0 = 1
> >a^(b+1) = a^b * a
> >a^b = sup {a^c : c < b}, if b is limit (1)
> >
> >Now this yields
> >
> >0^0 = 1
> >0^n = 0 for 0 < n < omega
> >
>
> Huh - very interesting.
>
0^(omega+1) < 0^omega
Interesting.
xi is successor ordinal iff 0^xi = 0
Also
xi is limit ordinal or 0 iff 0^xi = 1
Result
nonzero xi compact iff 0^xi = 0
Philosophical
nonzero xi not compact iff 0^xi = 1
> >I think, that (1) should be corrected to
> >a^b = sup {a^c : 0 < c < b}, if b is limit
>
Alan Sagan
This statement is incorrect: a^0=1 where a IS NOT 0 because 0^0 is undefined.
> a^(b+1) = a^b * a
> a^b = sup {a^c : c < b}, if b is limit (1)
> Now this yields
> 0^0 = 1
Stephen J. Herschkorn
Gerald wrote:
I have noticed this inconsistency as well. I agree with your
correction, and I have thought of it previously myself. Also, take a
look at Exercise 7 in Kunen; there he presents an allegedly equivalent
definition of the ordinal a^b as the type of a specific well-ordering
of the set of functions from b to a with finite support. Under this
defintion, 0^omega must be 0, since the set of such functions is empty,
Also, http://mathworld.wolfram.com/OrdinalExponentiation.html restricts
the definition to a != 0 for b a limit ordinal. It cites Rubin JE,
Set Theory for the Mathematician and Suppes, Axiomatic Set Theory for
this definition.
--
Stephen J. Herschkorn hers...@rutcor.rutgers.edu
Virgil
Dave Seaman
>> a^0 = 1
> This statement is incorrect: a^0=1 where a IS NOT 0 because 0^0 is undefined.
Your correction is incorrect; a^0 does indeed equal 0 for all a in
ordinal arithmetic, and also in cardinal arithmetic. In the latter case,
it's the cardinality of the set of mappings from the empty set into some
set A, which is 1.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
shedar
news:a6bf9368.0410...@posting.google.com...
On the Question of 0^0 (in ordinal exponentiation):
The question of whether 0^0 should be defined or left undefined is one of
those "immortal" questions in mathematics. The following sci.math FAQ by
Alopez gives a good account for both sides of the fence. While Cauchy
clearly treated it as undefined, the great Euler voted for "0^0=1". In more
recent times, Donald Knuth, for example, gives a pretty convincing argument
why 0^0 should be defined as 1 (by appealing to the beauty of the binomial
theorem). If usefulness and consistency are the yardstick in adopting a
convention, one should probably side with Euler and Knuth:
http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/
Should one vote for a prettier binomial theorem and for the fact that the
number of mappings from the empty set to the empty set is 0^0 (hence, "1"),
and live with the (unhappy) consequences of having discontinuities for the
function x^y? The answer is perhaps best left to the particular branch of
mathematics.
On the Question of 0^omega (in ordinal exponentiation):
> This would mean that the function 0^a is not continous at a = omega.
What harm is a little discontinuity at omega? It is already discontinuous at
zero (if one were to accept 0^0=1). I guess it means that many set theorists
have implicitly adopted the "somewhat arbitrary" convention that 0^0=1
(which, together with your equation (1), gives the *strange* result of
"0^omega=1"). But at least the "0^0"-part would make Euler and Knuth happy,
:-)
In view of the fact that 1^omega = 1 (according to the classical definition,
which also agrees with our intuition if we could call it *intuition* at
all), it only seems fair that one would desire 0^omega=0. So, I agree with
you to the extent that if one's goal were to make it come out as
"0^omega=0", then, the prevalent definition (via recursion) might need to be
modified as you have indicated.
In contrast--for your reference--see p. 204 in Jean E. Rubin's Set Theory
for the Mathematician (1967 Holden-Day). There, Rubin defines ordinal
exponentiation via "infinite products", and the following results are
DERIVED.
[Notation: Let On denote the class of all ordinals.]
Theorem 8.5.22 (p.204 J.E. Rubin, Set Theory for the Mathematician)
(a) For every a in On, a^0=1.
(b) For every a,b in On, a^(b+1) = (a^b) a.
(c) For every b in On, if b is a limit ordinal, then 0^b=0.
(d) For every a, b in On, if b is a limit and a =/= 0, then
a^b = Union{a^c : c in b}.
At least Jean E. Rubin's result in (c) above appears to agree with your
desire, and her result in (d) is not applicable when a = 0. [But I have to
admit that I have not gone through the details of developing ordinal
exponentiation via "infinite products", so I can't say if her definition of
ordinal exponentiation is equivalent to the prevalent version as you have
indicated above.]
Shedar
Eur Ing P.Stefanides
>Hi
>
>A question about ordinal exponentiation. In my opinion 0^a should be 0
>for every nonzero ordinal number a. But the recursive definition of
>ordinal exponentiation in all books and texts i found is stated as:
>
>a^0 = 1
>a^(b+1) = a^b * a
>a^b = sup {a^c : c < b}, if b is limit (1)
>
>Now this yields
>
>0^0 = 1
>0^n = 0 for 0 < n < omega
Adding to this my comments:
>From simple logarithmic consideration :
Log[0]X=0 for: 0 < X < + infinity
so 0^0= X
Log[0]0= I (I=INDETERMINATE ) ,for : - infinity < I <+ infinity ,
so 0^I=0
http://www.stefanides.gr/log_b(x).htm
Panagiotis Stefanides
David C. Ullrich
wrote:
Do you really think any actual set theorists would agree that 0^omega
"is really" equal to 1, instead of just saying that oops, the above
shows that a common definition of ordinal exponentiation isn't quite
right in a certain special case?
>In view of the fact that 1^omega = 1 (according to the classical definition,
>which also agrees with our intuition if we could call it *intuition* at
>all), it only seems fair that one would desire 0^omega=0. So, I agree with
>you to the extent that if one's goal were to make it come out as
>"0^omega=0", then, the prevalent definition (via recursion) might need to be
>modified as you have indicated.
>
>In contrast--for your reference--see p. 204 in Jean E. Rubin's Set Theory
>for the Mathematician (1967 Holden-Day). There, Rubin defines ordinal
>exponentiation via "infinite products", and the following results are
>DERIVED.
>
>[Notation: Let On denote the class of all ordinals.]
>Theorem 8.5.22 (p.204 J.E. Rubin, Set Theory for the Mathematician)
> (a) For every a in On, a^0=1.
> (b) For every a,b in On, a^(b+1) = (a^b) a.
> (c) For every b in On, if b is a limit ordinal, then 0^b=0.
> (d) For every a, b in On, if b is a limit and a =/= 0, then
> a^b = Union{a^c : c in b}.
>
>At least Jean E. Rubin's result in (c) above appears to agree with your
>desire, and her result in (d) is not applicable when a = 0. [But I have to
>admit that I have not gone through the details of developing ordinal
>exponentiation via "infinite products", so I can't say if her definition of
>ordinal exponentiation is equivalent to the prevalent version as you have
>indicated above.]
>
> Shedar
>
************************
David C. Ullrich
Dave Seaman
What do logarithms have to do with ordinals? What is the log of omega?
shedar
news:9uuvl0l74o6lnjjqv...@4ax.com...
>
> Do you really think any actual set theorists would agree that 0^omega
> "is really" equal to 1, instead of just saying that oops, the above
> shows that a common definition of ordinal exponentiation isn't quite
> right in a certain special case?
>
As I have indicated in the rest of my earlier post about J.E. Rubin's
formulation, I'm more inclined to agree with Gerald (the original poster)'s
suggestion that the prevalent definition probably has an "oops" in it,
especially when J.E. Rubin seems to agree that 0^omega (ordinal
exponentiation) should equal "0" (as indicated in her book). But I'd like to
know what set theorists think in case there is a deeper reason why one might
prefer it to equal to "1".
Any set theorists out there who can clarify?
Shedar
Stephen J. Herschkorn
>As I have indicated in the rest of my earlier post about J.E. Rubin's
>formulation, I'm more inclined to agree with Gerald (the original poster)'s
>suggestion that the prevalent definition probably has an "oops" in it,
>especially when J.E. Rubin seems to agree that 0^omega (ordinal
>exponentiation) should equal "0" (as indicated in her book). But I'd like to
>know what set theorists think in case there is a deeper reason why one might
>prefer it to equal to "1".
>
>Any set theorists out there who can clarify?
>
I am no set theorist; Fred Galvan and Herman Rubin (and perhaps Torkel
Franzen) seem to fit that characterization here. However, no one has
commented on a fact I posted earlier in this thread.
Kunen, who certainly is a set theorist, presents the definition
presented by the OP; one infers therefrom that 0^omega = 1. However,
right before presenting this recursive definition, Kunen refers to an
exercise where ordinal exponentiation a^b is defined directly as the
type of a certain well-ordering on a certain set of the functions from
b to a. (Kunen uses the Von Neumann definition of ordinal.) Since
this last set must be empty when a = 0 and b = omega, by the direct
definition, 0^omega = 0. I think Kunen would agree that there is a
glitch in the recursive definition.
For the interested reader, here is Kunen's "direct combinatorial
definition of a^b" (ordinal exponentiation). Let F be the set of
functions from b to a with finite support. For distinct f and g
in F, say f R g iff f(c) < g(c) where c = max{d in b: f(d) !=
g(d)}. Then a^b is the order type of (F, R).
Eur Ing Panagiotis Stefanides
>On Sun, 3 Oct 2004 12:39:58 +0000 (UTC), "Eur Ing P.Stefanides" wrote:
>> On 02 Oct 2004, Gerald wrote:
>>>Hi
>>>
>>>A question about ordinal exponentiation. In my opinion 0^a should be 0
>>>for every nonzero ordinal number a. But the recursive definition of
>>>ordinal exponentiation in all books and texts i found is stated as:
>>>
>>>a^0 = 1
>>>a^(b+1) = a^b * a
>>>a^b = sup {a^c : c < b}, if b is limit (1)
>>>
>>>Now this yields
>>>
>>>0^0 = 1
>>>0^n = 0 for 0 < n < omega
>
>> Adding to this my comments:
>
>>>From simple logarithmic consideration :
>> Log[0]X=0 for: 0 < X < + infinity
>> so 0^0= X
>
>> Log[0]0= I (I=INDETERMINATE ) ,for : - infinity < I <+ infinity ,
>>
>> so 0^I=0
>
>What do logarithms have to do with ordinals? What is the log of omega?
My simple interpretation of infinity,for "O" omega :
To zero base Log[+infinity]=I, [INDETERMINATE],also,
To +infinity base Log[+infinity]= I
Regards and my thanks
P.Stefanides
http://www.stefanides.gr/log_b(x).htm
>Dave Seaman
>Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
><<a href="http://www.commoncouragepress.com/index.cfm?action=book&bookid=228">http://www.commoncouragepress.com/index.cfm?action=book&bookid=228</a>>
Dave Seaman
> On 03 Oct 2004, Dave Seaman wrote:
>>>>From simple logarithmic consideration :
>>> Log[0]X=0 for: 0 < X < + infinity
>>> so 0^0= X
>>
>>> Log[0]0= I (I=INDETERMINATE ) ,for : - infinity < I <+ infinity ,
>>>
>>> so 0^I=0
>>
>>What do logarithms have to do with ordinals? What is the log of omega?
> I, anticipate them as numbers.
But not as ordinals, evidently.
> My simple interpretation of infinity,for "O" omega :
> To zero base Log[+infinity]=I, [INDETERMINATE],also,
> To +infinity base Log[+infinity]= I
But "+infinity" and "INDETERMINATE" are not ordinals.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>
David W. Cantrell
> But "+infinity" and "INDETERMINATE" are not ordinals.
Yeah, I knew all along that they were inordinate!
David C.
Todd Trimble
>On Sun, 3 Oct 2004 21:28:09 +0000 (UTC), Eur Ing Panagiotis Stefanides wrote:
>> On 03 Oct 2004, Dave Seaman wrote:
>>>>>From simple logarithmic consideration :
>>>> Log[0]X=0 for: 0 < X < + infinity
>>>> so 0^0= X
>>>
>>>> Log[0]0= I (I=INDETERMINATE ) ,for : - infinity < I <+ infinity ,
>>>>
>>>> so 0^I=0
>>>
>>>What do logarithms have to do with ordinals? What is the log of omega?
>> I, anticipate them as numbers.
>
>But not as ordinals, evidently.
>
And I anticipate that he wasn't talking about Conway numbers,
either. ;-)
>> My simple interpretation of infinity,for "O" omega :
>
>> To zero base Log[+infinity]=I, [INDETERMINATE],also,
>> To +infinity base Log[+infinity]= I
>