proof of theorems in additively indecomposable

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$\mathbb{H}$ is closed.
Let $$ be some increasing sequence of elements of $\mathbb{H}$ and let $$. Then for any $$, it must be that $$ and $$ for some $$. But then $$.

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$\mathbb{H}$ is unbounded^{}.
Consider any $\alpha $, and define a sequence^{} by ${\alpha}_{0}=S\alpha $ and ${\alpha}_{n+1}={\alpha}_{n}+{\alpha}_{n}$. Let $$ be the limit of this sequence. If $$ then it must be that $$ and $$ for some $$, and therefore $$. Note that ${\alpha}_{\omega}$ is, in fact, the next element of $\mathbb{H}$, since every element in the sequence is clearly additively decomposable.

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${f}_{\mathbb{H}}(\alpha )={\omega}^{\alpha}$.
Since $0$ is not in $\mathbb{H}$, we have ${f}_{\mathbb{H}}(0)=1$.
For any $\alpha +1$, we have ${f}_{\mathbb{H}}(\alpha +1)$ is the least additively indecomposable number greater than ${f}_{\mathbb{H}}(\alpha )$. Let ${\alpha}_{0}=S{f}_{\mathbb{H}}(\alpha )$ and ${\alpha}_{n+1}={\alpha}_{n}+{\alpha}_{n}={\alpha}_{n}\cdot 2$. Then $$. The limit case is trivial since $\mathbb{H}$ is closed and unbounded, so ${f}_{\mathbb{H}}$ is continuous.
Title  proof of theorems in additively indecomposable 

Canonical name  ProofOfTheoremsInAdditivelyIndecomposable 
Date of creation  20130322 13:29:07 
Last modified on  20130322 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 