# Ordinal exponentiation

• September 22nd 2010, 11:50 PM
sroberts
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
• September 23rd 2010, 04:46 AM
emakarov
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.
Quote:

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., $\emptyset$.
• September 23rd 2010, 10:40 AM
MoeBlee
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.]
• September 23rd 2010, 10:49 AM
emakarov
Well, doesn't the OP show that this definition is incorrect?

The MathWorld cites two books.
• September 23rd 2010, 10:52 AM
MoeBlee
Quote:

Originally Posted by sroberts
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}.]
• September 23rd 2010, 10:53 AM
MoeBlee
No. The poster's calculation is incorrect.

[EDIT: I'm mistaken.]
• September 23rd 2010, 12:03 PM
emakarov
But 0^0 = 1 both from the definition by recursion and because there exists a single, namely, empty, function from 0 to 0. Therefore, $\bigcup_{b<\omega} 0^b\supseteq 1$.
• September 23rd 2010, 12:13 PM
sroberts
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
• September 23rd 2010, 01:11 PM
MoeBlee

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.)
• September 23rd 2010, 03:02 PM
sroberts
Hi,

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

Thanks for the help from both of you.

Cheers
Sam
• September 24th 2010, 07:22 AM
MoeBlee
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.
• September 24th 2010, 07:32 AM
sroberts
That makes a lot more sense.

Thanks again,
Sam