Definition[]
O[k] is the least order-k number. Orders are defined as such: The first ordinal of order n is the first number such that it surpasses all other numbers in the previous order. Order 0 are the finite numbers, Order 1 are the infinite numbers, Order 2 are the "strange infinite" numbers, and so on.
Numbers[]
- O[0] = 0
- O[1] = Omega (ω)
- O[2] = Absolute Infinity (Ω)
- O[3] = Absolute True End (ↂ)
- O[4] = ????
- etc.
Definition: Order[]
Order is the least non-zero order fixed point. That is, the first number, a, greater than 0, such that: O[a] = a. This number can be visualized as O[O[O[....]]]. It may also be represented simply by O[0,1]. This obviously implies that this idea can be extended much much further.
Size[]
Unfortunately, this definition is not precise enough to call it well defined. This is because while O[a] is the least "ath-order number" by definition, order is not actually defined, except by listing out the first 4 examples. The idea is the property in question is to be inferred from a finite number of examples. However this is not rigorous enough to say what happens at 0[4]. All we can say, is by definition, this number is greater than Absolute True End, since this is O[3], and it is implied that O[a] < O[b] for a < b (that is it's monotonically increasing). The least non-zero fixed point (note that 0 is a fixed point since O[0] = 0} must be larger than any non-zero finite number, since 0[n] >= w for n > 0, which means O[n] >= w > n --> O[n] > n, and so can not be a fixed-point. In fact all transfinite ordinals are below O[2], so any transfinite ordinal a, will yield, O[a] > a. It can be shown to hold for an incredibly long time that O[x] > x. However eventually we must reach a fixed point when we get to the first infinite nesting of the O function it self. Let O equal the limit of O[1], O[O[1]] , O[O[O[1]]], ... etc. O[O] has the limit, O[O[1]] , O[O[O[1]]], O[O[O[O[1]]]], ... etc. Despite the offset ... it still has the same limit. Hence O is a fixed point, and in fact is the least such non-zero number.
The problem is that we have no way to determine how large the gaps between O[x] and O[x+1] are. Therefore we can not say how large Order is. Since it's at least larger than ATE, and ATE is Class 5 however, we know Order is at least a Class 5 number. The problem is it could be so large it's in Class 6 or even Dhaxarum, the Final Class. Ironically this number was placed in Class 3 for a while, simply because Class 3 was deemed to have too few members. This was added to pad that out. However it's very clear from it's definition that its larger than ATE, whatever it actually is, and this leads to a paradox placing it in Class 3, since ATE is the least member of Class 5.
It can however be argued that Order's correct placement is exactly Class 5, and not 6 or Dhaxarum. Why? Firstly, there is another number, Absolute Qalandar, that clearly is going for a similar idea of placing numbers into "orders" of some type. This suggests this number belongs in the same Class as Absolute Qalandar. Secondly Class 6 introduces the idea of dimensions. The clear assumption here is that Orders from a "straight continuous line" of numbers. There is no idea of it skipping to other dimensions. For this to be the case it would have to be the case that the entire "1-dimensional numberline" was some Order O[x], below Order. This seems unlikely. In any case it would still not be able to reach Dhaxarum. Why? Because the least member is Outerconst. Anything which could be explained with only A-mathematics is still in A-mathematics. The idea of fixed-points is still an A-mathematics concept. This means all such extensions of O[x] can never pass Outerconst.
In any case, until there is a definite theory about dimensions, we can not say for sure whether this is Class 5 or 6, but it is conjectured to lie in 5.