Skip to main content
added 56 characters in body
Source Link
JSCB
  • 1.6k
  • 10
  • 20

Denote the class of surreal numbers No. We can create new "number", like the gap $\infty=\{\infty^L|\infty^R\}$, defined by $\infty^L=\{x:\exists n\in\mathbb N,x<n\}$ and $\infty^R=\{x:\forall n\in\mathbb N,x>n\}$. $\infty\notin$No because $\infty^L$ and $\infty^R$ are not sets, but classes. Apparently we can construct new numbers in the form $\{L|R\}$, where $L$ and $R$ are subclasses of No such that no $x\in L$ is $\geq$ any $y\in R$. Then we can use these new numbers to construct more numbers, and then continue to expand No (possible indefinitely?). Is there any reason we do not do so?

Denote the class of surreal numbers No. We can create new "number", like the gap $\infty=\{\infty^L|\infty^R\}$, defined by $\infty^L=\{x:\exists n\in\mathbb N,x<n\}$ and $\infty^R=\{x:\forall n\in\mathbb N,x>n\}$. $\infty\notin$No because $\infty^L$ and $\infty^R$ are not sets, but classes. Apparently we can construct new numbers in the form $\{L|R\}$, where $L$ and $R$ are subclasses of No such that no $x\in L$ is $\geq$ any $y\in R$, and then continue to expand No (possible indefinitely?). Is there any reason we do not do so?

Denote the class of surreal numbers No. We can create new "number", like the gap $\infty=\{\infty^L|\infty^R\}$, defined by $\infty^L=\{x:\exists n\in\mathbb N,x<n\}$ and $\infty^R=\{x:\forall n\in\mathbb N,x>n\}$. $\infty\notin$No because $\infty^L$ and $\infty^R$ are not sets, but classes. Apparently we can construct new numbers in the form $\{L|R\}$, where $L$ and $R$ are subclasses of No such that no $x\in L$ is $\geq$ any $y\in R$. Then we can use these new numbers to construct more numbers, and continue to expand No (possible indefinitely?). Is there any reason we do not do so?

Source Link
JSCB
  • 1.6k
  • 10
  • 20

Going beyond the surreal numbers

Denote the class of surreal numbers No. We can create new "number", like the gap $\infty=\{\infty^L|\infty^R\}$, defined by $\infty^L=\{x:\exists n\in\mathbb N,x<n\}$ and $\infty^R=\{x:\forall n\in\mathbb N,x>n\}$. $\infty\notin$No because $\infty^L$ and $\infty^R$ are not sets, but classes. Apparently we can construct new numbers in the form $\{L|R\}$, where $L$ and $R$ are subclasses of No such that no $x\in L$ is $\geq$ any $y\in R$, and then continue to expand No (possible indefinitely?). Is there any reason we do not do so?