User:Vel!/Surreal numbers


 * This page is for open editing. You are free to modify, add, expand, or discuss.

This page is about surreal numbers, and specifically their relation to googology. If you're looking Surreal Analysis 101, go look it up or something &mdash; we won't replicate an entire intro for you here.

Fundamental sequences
The notion of the fundamental sequence (FS) has to be modified to be meaningful in On2. Vanilla ordinals can be considered as a limit of a certain monotonic sequence which serves as the FS. Over surreals, however, FS's may be non-monotonic (such as 0, w, 1, w-1, 2, w-2, ... being an FS for w/2), and in fact they must be that way in almost all cases. In addition, we need to define our own way of determining the "limit" of that sequence.

Let S be a countably infinite set of surreal numbers. (That condition is important in the uniqueness proof.) Partition S into two disjoint sets L and R so that:


 * 1) L is well-ordered by <.
 * 2) L has order type of most w with respect to <.
 * 3) R is well-ordered by >.
 * 4) R has order type at most w with respect to >.
 * 5) All elements in L are strictly less than all those in R.

Then lim2(S) = {L|R}. This is not a total function, but it is single-valued. We can prove this by the uniqueness of the partition:


 * Let L1/R1 and L2/R2 be two such partitions of S. We wish to show that L1 = L2, which proves the uniqueness of the partition.
 * If L1 and L2 are both empty, then we are done. Otherwise, let x be the least element of L1, which exists because L1 is well-ordered by <. x must therefore be the least element of S (any smaller elements must belong to R1, which contradicts condition 3). Now x belongs to either L2 or R2. Suppose x belongs to R2: L2 is empty and x is the least element of R2. But since S is infinite, so is R2, and therefore its order type with respect to < is at least w. Since R2 has a least element, its order type cannot be w, which contradicts condition 2. Therefore, x is in L2.
 * Since x is in both L1 and L2, we consider S \ {x} as partitioned by (L1 \ {x})/R1 and (L2 \ {x})/R2. We can make the same argument by induction, and induction succeeds because L is well-ordered.