proof of theorems in additively indecomposable

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$\mathbb{H}$ is closed.

Let $\{\alpha_{i}\mid i<\kappa\}$ be some increasing sequence of elements of $\mathbb{H}$ and let $\alpha=\sup\{\alpha_{i}\mid i<\kappa\}$ . Then for any $x,y<\alpha$ , it must be that $x<\alpha_{i}$ and $y<\alpha_{j}$ for some $i,j<\kappa$ . But then $x+y<\alpha_{\max\{i,j\}}<\alpha$ .
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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 $\alpha_{\omega}=\sup_{n<\omega}\alpha_{n}$ be the limit of this sequence. If $x,y<\alpha_{\omega}$ then it must be that $x<\alpha_{i}$ and $y<\alpha_{j}$ for some $i,j<\omega$ , and therefore $x+y<\alpha_{\max\{i,j\}+1}$ . 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 $\alpha_{0}=Sf_{\mathbb{H}}(\alpha)$ and $\alpha_{n+1}=\alpha_{n}+\alpha_{n}=\alpha_{n}\cdot 2$ . Then $f_{\mathbb{H}}(\alpha+1)=\sup_{n<\omega}\alpha_{n}=\sup_{n<\omega}S\alpha\cdot 2% \cdots 2=f_{\mathbb{H}}(\alpha)\cdot\omega$ . 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	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