Infiniter?

[Kuroneko]

Okay. I've only worked with Euclidean spaces thus far in my education, so I see now where my (admittedly yet small) education lacks.

Hmm... a measure on a space assigns numbers (not necessarily real and often infinite) to subsets of that set subject to some constraints that measures the 'content' o that subset (length, area, volume, ...). The most common measure is called the Lebesgue measure; this is what you've been using all along, even if it was never made explicit, which is generated by open balls in Euclidean space. Look up the Banach-Tarski theorem, which states that a ball can be partitioned into finitely many pieces and rearranged by rigid motions into two balls of the same volume. Interepreted naively, it sounds as if 1 = P1+P2+P3+...+Pn = 2. It is not a contradiction because the pieces involved are not measureable, and thus do not have volume (note that zero measure and infinite measure are not the same thing as being non-measurable; for non-measurable sets in three dimensions, the concept of volume is not definable at all).

0?

If x is the variable of integration, [d/dx]^{1/2}[c] = c/[sqrt(πx)]. If you're interested, some time ago I found a pretty good introduction as to the meaning of fractional derivatives and integrals as a paper on arxiv.org; I can probably find it again.

[Ariphaos]

Well, that depends. The aleph concepts are 19th-century, although it wasn't until the 20th that they have received extensive study.

Well, we were introduced to analysis as being 19th century math, did the basic countable-uncountable and set theory items but nowhere near what you speak of.

Well, let's start with something relatively simple--ordinal numbers, which every student of set theory should be familiar with (they have a very intimate connection with the axiom of choice). A set α is an ordinal iff every element β∈α is a subset of α and α is well-ordered by ∈ (a 'well-order' means that any collection of ordinals contains a smallest element, and ∈ is what is used to compare them). That's probably better appreciated by example. Define 0 = ∅ (empty set), 1 = {0} = {∅}, 2 = {0,1} = {∅,{∅}}, 3 = {0,1,2} = {∅,{∅},{∅,{∅}}}, and in general n+1 = n∪{n}. Note that β<α means that α contains β (β∈α). Ordinals have a number of interesting properties. For example, given a set of ordinals Α, the smallest ordinal greater than or equal to all numbers in Α ("supremum", or "sup Α") is the union of all the ordinals in Α (e.g., 1∪2∪3 = 3), and the smallest ordinal contained in it is the intersection (e.g., 2∩3∩4 = 2).

Arithmetic can be defined recursively: α+0 = α, α+1 = α∪{α}, and α+β as the supremum of all {(α+γ)+1: γ<β}. For example, 2+1 = 2∪{2} = {0,1,2} = 3, and 2+2 = sup{(2+0)+1,(2+1)+1} = sup{3,4} = 4, etc. Multiplication is done likewise: α0 = 0 and αβ = sup{αγ+α: γ<β}, e.g., α1 = α0+α = α, α2 = sup{α0+α,α1+α} = α+α, just as expected. Exponentiation: α^0 = 1, α^β = sup{(α^γ)α: γ<β}, e.g., α^1 = sup{(α^0)α} = α, α^2 = sup{(α^0)α, (α^1)α} = αα.

Infinite ordinal numbers come in when one realizes that there no reason to stop with the finite ones. The smallest infinite ordinal is ω = {0,1,2,3,...}. Immediately, ω+1 = {0,1,2,3,...,ω} that is distinct from ω, and so on for ω+n. The ordinal ω2 is the smallest ordinal greater than any of {ω+n}, ω² is the smallest ordinal greater than any of {ωn}, ω³ is the smallest ordinal greater than any of {ω²+n}, and so on to ω^ω and beyond. As an aside, in ZFC it is possible to define a 'cardinal number' as an 'ordinal number not equinumerous to any lesser ordinal'. They have a different arithmetic, however.

I feel like this kind of thing cheats God and I don't even believe in one.

Exercise 1: Does 1+ω = ω+1 and does 2ω = ω2?

This looks so suspiciously like a trap. {ω, 0, 1, 2, 3, ...} should be equal to {0, 1, 2, 3, ..., ω} since ordering them is for convenience (if I'm understanding this right). Also the union operator functions so that a U b = b U a.

It seems though you want me to think of the chain that occurs in 1 + ω. We get:

1 + ω = sup {(1 + ω-1)+1, (1 + ω-2)+1, (1 + ω-3)+1, ... (1 + 0) + 1}

Right? It doesn't seem to change the answer.

Going with that, the same would hold true for 2ω = ω2.

2ω = sup {2*0+2, 2*1+2, ... 2*(ω-1)+2}

Though this is a bit harder for me to wrap my head around.

Exercise 2: Show that the ordinals do not form a set.

I can show a set that's not an ordinal, does that count? :-p

Err, this seems rather hard, since every single ordinal is a proper subset of another ordinal, namely, ordinal n is a subset of n+1 for any n. Is it impossible to truly encapsulate the set? That is, since ω is a valid ordinal, there is no equation which can represent every ordinal, since ω^ω^ω...^ω is still a valid ordinal and no matter how silly we get there will be another ordinal after it?

Surreal numbers are a very rich field; I can introduce them if you like, but be aware that they are probably too diffucult to get accustomed to in one sitting. The surreal numbers are nicer than ordinals because they form a field (i.e., their arithmetic operations preserve all of the properties of real numbers, such as having multiplicative inverses, which ordinals and cardinals lack).

Possibly, I have way too many interests though, myself. Assuming my above answers are correct being able to do algebra with infinity (so to speak) will do for math for now :-)

If you find that counterintuitive, well, there's also fractional calculus (quick, what's the 1/2th derivative of a constant?).

Its square root? Just a wild guess.

[Ford Prefect]

Cheers for that everyone, he get's it, though still had innate problems with the concept of 'more infinite'. He was mostly arguing from "the intuitive notion of 'infinite'" as Surly put it. And I have to say, it makes more sense to me now as well.

[Ariphaos]

If x is the variable of integration, [d/dx]^{1/2}[c] = c/[sqrt(πx)]. If you're interested, some time ago I found a pretty good introduction as to the meaning of fractional derivatives and integrals as a paper on arxiv.org; I can probably find it again.

[Homer]DOH![/homer]

It actually seems pretty straightforward compared to, say, differential equations.

[Ariphaos]

Cheers for that everyone, he get's it, though still had innate problems with the concept of 'more infinite'. He was mostly arguing from "the intuitive notion of 'infinite'" as Surly put it. And I have to say, it makes more sense to me now as well.

I think a better way of describing the mathematical difference is to go back to Surly's original post in this thread.

You have integers. You can count them. Zero, one, two, three, four, five, six... until eternity ends several times over. You still won't be done, but you counted a few of them in an orderly fashion.

This type of infinity is called countably infinite. Strangely enough.

Now, let's have a race! Start counting the real numbers!

...

With rational numbers, you can cheat (see the bijection in Surle's original post again) - but this is not possible with real numbers. There's no place to go after 'zero'. The number of zeros before the first one is a countably infinite number, but beyond that, you're screwed.

Thus, real numbers are called 'uncountable'.

[Kuroneko]

Well, we were introduced to analysis as being 19th century math, did the basic countable-uncountable and set theory items but nowhere near what you speak of.

Analysis is 'mostly' 19th-century, although it has had much conceptual development throughout the ages, at least to Archimedes. But even Cantor's set theory already had the notion of "order type", which is completely equivalent to the more modern ordinal number treatment. This is really classic mathematics, not new research.

Exercise 1: Does 1+ω = ω+1 and does 2ω = ω2?

This looks so suspiciously like a trap. ... It seems though you want me to think of the chain that occurs in 1 + ω. We get:
1 + ω = sup {(1 + ω-1)+1, (1 + ω-2)+1, (1 + ω-3)+1, ... (1 + 0) + 1}
Right? It doesn't seem to change the answer.

But ω is the smallest ordinal greater than all the finite ordinals {0,1,2,...}. Clearly, it does not admit a predecessor ω-1, since such a thing would be greater than any finite ordinal but lesser than ω. Try that again.

Going with that, the same would hold true for 2ω = ω2. 2ω = sup {2*0+2, 2*1+2, ... 2*(ω-1)+2}

Same problem.

Exercise 2: Show that the ordinals do not form a set.

I can show a set that's not an ordinal, does that count? :-p Err, this seems rather hard, since every single ordinal is a proper subset of another ordinal, namely, ordinal n is a subset of n+1 for any n.

True. And that's an important property.

Is it impossible to truly encapsulate the set? That is, since ω is a valid ordinal, there is no equation which can represent every ordinal, since ω^ω^ω...^ω is still a valid ordinal and no matter how silly we get there will be another ordinal after it?

That's true; even after ε_0 = ω^ω^ω^... (non-terminating; the least ordinal greater than all of {ω,ω^ω,ω^ω^ω,...}) one can do ε_1 = (ε_0)^(ε_0)^..., and even more silliness. But don't think in terms of equations--if there is a set of all ordinals, what must follow?

Its square root? Just a wild guess.

Good guess. Some multiple of 1/sqrt(x), actually, if x is the variable of differentiation. Intuitively, it makes a sense--nth-derivative decreases power by n; 1/2-th derivative decreases power by 1/2. The extra 1/sqrt(π) comes from a generalization of the factorial (cf. gamma function), since nth-derivatives of x^m introduce a factor of m!/(m-n)!.

It actually seems pretty straightforward compared to, say, differential equations.

It's can be treated as an integral of a related function, of actually, as well as having more geometrical interpretations involving projections.

[Kuroneko]

Now, let's have a race! Start counting the real numbers! ...
There's no place to go after 'zero'. The number of zeros before the first one is a countably infinite number, but beyond that, you're screwed. Thus, real numbers are called 'uncountable'.

It's a fairly good explanation, but there are some caveats to the above excerpt relating to the current tangent in this thread. The axiom of choice is equivalent to the statement that every set X has a bijection to some ordinal α. Thus, there is a 0th, 1st, 2nd, ..., element, etc. The difference between countable and uncountable sets is that uncountable sets will always have an ωth, (ω+1)st, ... (ω2)nd, ..., ω²th, ..., etc., at least through all of the countable ordinals. For an arbitrary αth real number, there will be an (α+1)st real number to go to. So, while the reals are not countable, they are well-orderable--there will be a "place to go" after any particular real number. "How many?" and "in what order?" are two separate questions, however, which is why cardinal and ordinal numbers (respectively answering those two questions) have different arithmetic.

[Ariphaos]

Gah, accidentally closed the window. Anyway, easy answer first.

That's true; even after ε_0 = ω^ω^ω^... (non-terminating; the least ordinal greater than all of {ω,ω^ω,ω^ω^ω,...}) one can do ε_1 = (ε_0)^(ε_0)^..., and even more silliness. But don't think in terms of equations--if there is a set of all ordinals, what must follow?

Call the set t. It would contain all ordinalls from the null set ('0'), to t-1. By definition, this makes t an ordinal outside the set it is supposed to contain. Contradiction.

Exercise 1: Does 1+ω = ω+1 and does 2ω = ω2?

This looks so suspiciously like a trap. ... It seems though you want me to think of the chain that occurs in 1 + ω. We get:
1 + ω = sup {(1 + ω-1)+1, (1 + ω-2)+1, (1 + ω-3)+1, ... (1 + 0) + 1}
Right? It doesn't seem to change the answer.

But ω is the smallest ordinal greater than all the finite ordinals {0,1,2,...}. Clearly, it does not admit a predecessor ω-1, since such a thing would be greater than any finite ordinal but lesser than ω. Try that again.

Alright, it seems we can still legally break down ω into 1 + 1 + 1 + 1 ... + 1. If that is true, the originating 1 + ... adds no real amount, and thus 1 + ω = ω?

Going with that, the same would hold true for 2ω = ω2. 2ω = sup {2*0+2, 2*1+2, ... 2*(ω-1)+2}

Same problem.

So, breaking ω up like that again, and multiplying 2 throughout, we would get a similar result - 2 + 2 + 2 + 2 + 2 ... + 2 = ω?

[NecronLord]

Exercise 1: Does 1+ω = ω+1 and does 2ω = ω2?
Exercise 2: Show that the ordinals do not form a set.

Just out of curiosity, are you a teacher Kuroneko?

Anyway, thing to remember when talking about infinities is that one's intuitive conception of infinity is the set of all integers from 1 upwards.

Hence, 'Emperor Palpatine claims infinite power' would not be interpreted, by most people, at least, as 'Emperor Palpatine claims that his power can be measured to any arbitrary degree of accuracy between 0 and 1,' but rather, with the above, more 'everyday' definition.

[drachefly]

[the idea of a fractional derivative] actually seems pretty straightforward compared to, say, differential equations.

It's can be treated as an integral of a related function, of actually, as well as having more geometrical interpretations involving projections.

I usually think of it in terms of the Fourier transform and properties around the derivative. Others might find this viewpoint useful too.

[Ariphaos]

Just out of curiosity, are you a teacher Kuroneko?

Either that or he wants to know I'm not wasting his time, or both.

Anyway, thing to remember when talking about infinities is that one's intuitive conception of infinity is the set of all integers from 1 upwards.

Err, most mathematicians I know consider that type of thing to be undefined, as in, 0/0, and that infinity is the opposite of zero.

[Kuroneko]

Call the set t. It would contain all ordinalls from the null set ('0'), to t-1. By definition, this makes t an ordinal outside the set it is supposed to contain. Contradiction.

You're basically correct, although the "t-1" is not guaranteed. Some ordinals do not have a predecessor, e.g., the only ordinals less than ω are the finite ones: {0,1,2,...}. There are no such things as "ω-1" in the ordinals. Try it--if there was, recall that ordinals are sets, so you should be able to specify its members; in particular if α=ω-1, or α+1=α∪{α} = ω, has no solution. Ordinals that have no predecessor are called "limit ordinals," appropriate.

Alright, it seems we can still legally break down ω into 1 + 1 + 1 + 1 ... + 1. If that is true, the originating 1 + ... adds no real amount, and thus 1 + ω = ω?

But there is no independent meaning of "sum of ω 1's" other than ω1, so that your argument isn't precise enough. The sum of ordinals was defined in terms of sums lower ordinals, but it's not really 'recursive' in the normal sense, since it may happen (as it does now) that the ordinals involved have no predecessor. See what happens in the supremum definition without the (ω-1)s, (ω-2)s, etc. (finite only).

Just out of curiosity, are you a teacher Kuroneko?

Guilty.

Anyway, thing to remember when talking about infinities is that one's intuitive conception of infinity is the set of all integers from 1 upwards. Hence, 'Emperor Palpatine claims infinite power' would not be interpreted, by most people, at least, as 'Emperor Palpatine claims that his power can be measured to any arbitrary degree of accuracy between 0 and 1,' but rather, with the above, more 'everyday' definition.

However, in terms of sets, "finite" or "infinite" has to do with how many things there are in that set. Even intuitively, this notion of "how many?" has nothing to do with the structure of the things discussed, so that whether or not they have a "beginning" or an "end" is simply not relevant. That's why Ford Prefect's friend was incorrect.

I usually think of it in terms of the Fourier transform and properties around the derivative. Others might find this viewpoint useful too.

That's actually a pretty neat viewpoint. (I wonder if the same thing could be done with Laplace transform if the summation is generalized in some way (Eucler-Maclaurin might work), although that seems too ugly compared to Fourier.) An integral of f(t) is usually thought of as an area between the x-axis and f(t), with some qualifiers (if f(t)<0 on a region, the area contribution is negative there). The fractional integrals have a likewise intepretation, with f(t) making a curve in plane three-dimensional space and then taking an area of a projection. If someone is interested, math.CA/0110241 contains mathematical and physical interpretations of the fractional integrals (although personally I think the physical interpretation provided is not adequate).

[Ariphaos]

You're basically correct, although the "t-1" is not guaranteed. Some ordinals do not have a predecessor, e.g., the only ordinals less than ω are the finite ones: {0,1,2,...}. There are no such things as "ω-1" in the ordinals. Try it--if there was, recall that ordinals are sets, so you should be able to specify its members; in particular if α=ω-1, or α+1=α∪{α} = ω, has no solution. Ordinals that have no predecessor are called "limit ordinals," appropriate.

Details always get me. :-/

But there is no independent meaning of "sum of ω 1's" other than ω1, so that your argument isn't precise enough. The sum of ordinals was defined in terms of sums lower ordinals, but it's not really 'recursive' in the normal sense, since it may happen (as it does now) that the ordinals involved have no predecessor. See what happens in the supremum definition without the (ω-1)s, (ω-2)s, etc. (finite only).

Okay, so, recursion cannot apply to the limit ordinals (null, ω, and other stuff?)

Since 1+9 still seems to equal 10 (sup {2, 3, 4, 5, 6, 7, 8, 9, 10} ) we get

1 + ω = sup {2, 3, 4, 5 ...} = ω

Are all other limit ordinals derived from ω?

It would seem you could introduce infinitesimals to this system, too?

[Kuroneko]

Okay, so, recursion cannot apply to the limit ordinals (null, ω, and other stuff?) Since 1+9 still seems to equal 10 (sup {2, 3, 4, 5, 6, 7, 8, 9, 10} ) we get 1 + ω = sup {2, 3, 4, 5 ...} = ω.

Exactly correct. Note that ω+1 ≠ ω+1 = ω. Ordinal arithmetic is not in general commutative.

Are all other limit ordinals derived from ω?

Not really, no. Any expression specifiying a limit ordinal would have to depend on ω somehow, but there are ordinals for which there are no expression. That is necessary--there are more ordinals than can "fit into" a set, and any schema for generating ordinals by expressions we could make would necessarily make only countably many (since we would have only countably may symbols and operations). In other words, there are always ordinals which one can't name no matter the language one invents to name them.

It would seem you could introduce infinitesimals to this system, too?

Not really--ordinal numbers are used for determining position, e.g., "3rd" or "100th" or "ωth" [1]. They simply don't measure quantity. If you're wondering how ""ωth place" could make sense, consider the following situation: it is logically (just not physiologically or physically) possible to perform an action in ever-decreasing time intervals. For example, 0th action taking 1/2, 1st 1/4, 2nd 1/8... what happens after 1 time unit has passed and we still want to perform that action again? We're on the "ωth" action, the smallest position after all of {0th,1st,2nd,...}.

Now, in surreal numbers, it is quite possible to have infinitesimals. Surreal numbers are defined recursively (not in the comp. sci. sense, though). The rule is simple: given any two sets of surreals A and B such that every surreal in A is less than all surreals in B, x = {A|B} is a surreal number. Since we don't have any surreals yet, the only thing we can do is use the empty set (which vacuously satisfies the above condition) to make 0 = {|}. From then on, we can make 1 = {0|}, 1/2 = {0|1}, 1/4 = {0|1/2}, 2 = {1|} = {0,1|}, ..., ω = {0,1,2,...|} [2], ι = {0|1,1/2,1/4,1/8,...}. The arithmetic and comparison operations on surreals are a bit complicated, so I won't go into them right now, but ι happens to be greater than 0 and less than any positive real. Of course, one can also have ι/n, ι², ... even ι^ω and even lesser positive surreals.

[1] Except that they start at "0th". There's an apocryphal story about Sierpinski losing a piece of baggage because he counted them as "zero, one, two...".
[2] This is different from the ordinal ω.

[Ariphaos]

Not really, no. Any expression specifiying a limit ordinal would have to depend on ω somehow, but there are ordinals for which there are no expression. That is necessary--there are more ordinals than can "fit into" a set, and any schema for generating ordinals by expressions we could make would necessarily make only countably many (since we would have only countably may symbols and operations). In other words, there are always ordinals which one can't name no matter the language one invents to name them.

Can that kind of thing be represented at all? Or is it just a required artifact of the math?

A better way of saying what I'm asking may be 'where does this pop up'?

Not really--ordinal numbers are used for determining position, e.g., "3rd" or "100th" or "ωth" [1]. They simply don't measure quantity. If you're wondering how ""ωth place" could make sense, consider the following situation: it is logically (just not physiologically or physically) possible to perform an action in ever-decreasing time intervals. For example, 0th action taking 1/2, 1st 1/4, 2nd 1/8... what happens after 1 time unit has passed and we still want to perform that action again? We're on the "ωth" action, the smallest position after all of {0th,1st,2nd,...}.

Well the usefulness of your particular example seems limited since there doesn't seem to be such a thing as division in ordinals, but I'm familiar with that kind of bounded infinity. I suppose that's what surreal numbers are for?

Now, in surreal numbers, it is quite possible to have infinitesimals. Surreal numbers are defined recursively (not in the comp. sci. sense, though). The rule is simple: given any two sets of surreals A and B such that every surreal in A is less than all surreals in B, x = {A|B} is a surreal number. Since we don't have any surreals yet, the only thing we can do is use the empty set (which vacuously satisfies the above condition) to make 0 = {|}. From then on, we can make 1 = {0|}, 1/2 = {0|1}, 1/4 = {0|1/2}, 2 = {1|} = {0,1|}, ..., ω = {0,1,2,...|} [2], ι = {0|1,1/2,1/4,1/8,...}. The arithmetic and comparison operations on surreals are a bit complicated, so I won't go into them right now, but ι happens to be greater than 0 and less than any positive real. Of course, one can also have ι/n, ι², ... even ι^ω and even lesser positive surreals.

It would seem like irrational numbers are a bit tricky here? Or do they work the same way?

[1] Except that they start at "0th". There's an apocryphal story about Sierpinski losing a piece of baggage because he counted them as "zero, one, two...".

Given the nature of my upbringing and education, I can relate.

[Kuroneko]

Can that kind of thing be represented at all? Or is it just a required artifact of the math?

Different representation schemas may represent completely different ordinals, so it may be the case that any particular ordinal could be named in some language, but the point is that no language can capture them all. (Language in the formal sense, with a finite alphabet--but it's actually worse, as long as the alphabet of the language forms a set, some ordinals would still slip through.)

A better way of saying what I'm asking may be 'where does this pop up'?

Inaccessible ordinals (and inaccessible cardinals) are almost their own field of study (some set theorists are concerned specifically with properties of large sets). Additionally...

Well the usefulness of your particular example seems limited since there doesn't seem to be such a thing as division in ordinals, but I'm familiar with that kind of bounded infinity. I suppose that's what surreal numbers are for?

... transfinite induction pops up when large well-ordered sets are studied, which carries induction through all the ordinals regardless of rank. An example relevant here would be proving certain properties of surreal numbers. Actually, transfinite induction through all of the ordinals is required to construct all the surreals in the first place. Ordinal number have their own very impressive mathematical applications; they just aren't meant to do what you have in mind.

It would seem like irrational numbers are a bit tricky here? Or do they work the same way?

They work the same way, but they require infinitely many numbers to be specified. Actually, any real number not in the form n+m/2^k for integer n,m,k requires infinitely many elements in the set-pair representation. The surreal notation x = {A|B} picks out the "simplest" number number between those in A and those in B, and for finitely many finite numbers in A and B, the "simplest" means "the minimal number with minimal k" because of the way arithmetic is defined. This gives a very natural notion of "simplicity" of numbers as which iteration of the creation rule is the earliest point at which the number is created. Zero has simplicity 0 since it is not dependent on other surreals; 1 and -1 have simplicity 1, since they are created at the next iteration ({0|} and {|0}, respective), then come -2,-1/2,1/2,2 ({|-1},{-1|0},{0|1},{1|}), and so on. All reals not in the form n+m/2^k (as well as ±ω and ±i = ±1/ω) are created on the ωth iteration. Then come ±(ω+1), ±(ω-1) and a host of others, and so on without bound.

[Ariphaos]

Different representation schemas may represent completely different ordinals, so it may be the case that any particular ordinal could be named in some language, but the point is that no language can capture them all. (Language in the formal sense, with a finite alphabet--but it's actually worse, as long as the alphabet of the language forms a set, some ordinals would still slip through.)

Would a schema that only allowed say, odd primes be a legal ordinal schema? (Since it would appear that no arithmatic could be used on it).

Inaccessible ordinals (and inaccessible cardinals) are almost their own field of study (some set theorists are concerned specifically with properties of large sets). Additionally...

Any place you recommend looking into this?

... transfinite induction pops up when large well-ordered sets are studied, which carries induction through all the ordinals regardless of rank. An example relevant here would be proving certain properties of surreal numbers. Actually, transfinite induction through all of the ordinals is required to construct all the surreals in the first place. Ordinal number have their own very impressive mathematical applications; they just aren't meant to do what you have in mind.

I don't doubt that. In any case, thanks for the math lesson, much appreciated.

[Kuroneko]

Any place you recommend looking into this?

On inacessibility in set theory? I could, but I don't think it would be productive for you to go into it until you're more experienced in set theory. A more basic, yet nontrivial, book on cardinal an ordinal numbers that I liked was: Cardinal and Ordinal Numbers by Wacław Sierpiński. If by the time you go through more set theory you're still interested in inaccessibility, a good (and very demanding!) book is The Higher Infinite by Akihiro Kanamori.

[Kuroneko]

Forgot about this...

Would a schema that only allowed say, odd primes be a legal ordinal schema? (Since it would appear that no arithmatic could be used on it).

Yes (in the terms of formal logic, not comp. sci., in which `schema' has a different meaning). The property of being an odd prime is Turing-decidable, so there is a (recursive) language that names all odd primes and only odd primes.