proof of theorems in additively indecomposable

  • is closed.

    Let {αii<κ} be some increasing sequence of elements of and let α=sup{αii<κ}. Then for any x,y<α, it must be that x<αi and y<αj for some i,j<κ. But then x+y<αmax{i,j}<α.

  • is unboundedPlanetmathPlanetmath.

    Consider any α, and define a sequenceMathworldPlanetmath by α0=Sα and αn+1=αn+αn. Let αω=supn<ωαn be the limit of this sequence. If x,y<αω then it must be that x<αi and y<αj for some i,j<ω, and therefore x+y<αmax{i,j}+1. Note that αω is, in fact, the next element of , since every element in the sequence is clearly additively decomposable.

  • f(α)=ωα.

    Since 0 is not in , we have f(0)=1.

    For any α+1, we have f(α+1) is the least additively indecomposable number greater than f(α). Let α0=Sf(α) and αn+1=αn+αn=αn2. Then f(α+1)=supn<ωαn=supn<ωSα22=f(α)ω. The limit case is trivial since is closed and unbounded, so f is continuous.

Title proof of theorems in additively indecomposable
Canonical name ProofOfTheoremsInAdditivelyIndecomposable
Date of creation 2013-03-22 13:29:07
Last modified on 2013-03-22 13:29:07
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 9
Author mathcam (2727)
Entry type Proof
Classification msc 03E10
Classification msc 03F15