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 $f_{\\mathbb{H}}(\\alpha+1)=\\sup_{n<\\omega}\\alpha_{n}=\\sup_{n<\\omega}S\\alpha\\cdot2%\\cdots 2=f_{\\mathbb{H}}(\\alpha)\\cdot\\omega$ . The limit case is trivial since $\\mathbb{H}$ is closed and unbounded, so $f_{\\mathbb{H}}$ is continuous.

单词	ProofOfTheoremsInAdditivelyIndecomposable
释义	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 $f_{\\mathbb{H}}(\\alpha+1)=\\sup_{n<\\omega}\\alpha_{n}=\\sup_{n<\\omega}S\\alpha\\cdot2%\\cdots 2=f_{\\mathbb{H}}(\\alpha)\\cdot\\omega$ . The limit case is trivial since $\\mathbb{H}$ is closed and unbounded, so $f_{\\mathbb{H}}$ is continuous.
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