Let $\alpha \in P_R$ be a cut.
Since there exists a cut that is not $\{q\in Q\mid q Let $$\gamma= 0^* \cup \{0\} \cup \{q\in P_Q\mid\text{ there exists }r\in P_Q\text{ such that }r>q\text{ and }1/r \notin \alpha\}\;.$$ I have proved that $\alpha \gamma$ is a subset of $1^*$.
I dont't know how to prove $1^*$ is a subset of $\alpha \gamma$. Help
Multiplication inverse for dedekind cut
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elementary-set-theory
1 Answers
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We have $\alpha\gamma=0^*\cup\{0\}\cup\{st:0t(1/r\notin\alpha)\}$. For positive $t$ the condition that there is some $r>t$ such that $1/r\notin\alpha$ is equivalent to the condition that there is a positive $r<1/t$ such that $r\notin\alpha$, i.e., that $\alpha\subsetneqq(1/t)^*$.
Let $q\in 1^*$; clearly $q\in\alpha\gamma$ if $q\le 0$, so assume that $0 We know that $q=a/b$ for some positive integers $a$ and $b$ such that $a
Let $p=\dfrac{b-a}a$; then $q^{-n}=(1+p)^n$, and it’s an easy induction to show that $(1+p)^n\ge 1+np$. Since $\langle 1+np:n\in\omega\rangle$ is clearly unbounded, we’re done.
$p≦st$} – 2012-06-11