# proof of theorems in additively indecomposable

• $\mathbb{H}$ 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 $x+y<\alpha_{\max\{i,j\}}<\alpha$.

• $\mathbb{H}$ 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.

• $f_{\mathbb{H}}(\alpha)=\omega^{\alpha}$.

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 ProofOfTheoremsInAdditivelyIndecomposable 2013-03-22 13:29:07 2013-03-22 13:29:07 mathcam (2727) mathcam (2727) 9 mathcam (2727) Proof msc 03E10 msc 03F15