Find a (nonstandard) ordering of the set of natural numbers N such that there exists a
nonempty proper subset A of N and such that sup A exists in N, but sup A is not in A.
Let $\displaystyle B=\{1-1/n\mid n=1,2,\dots\}$ and let $\displaystyle M=B\cup\{1\}$. Then B is a nonempty proper subset of M and, with respect to the standard ordering, there exists sup B in M, but sup B is not in B. It is sufficient to define a one-to-one correspondence $\displaystyle f : M\to\mathbb{N}$ and consider the order carried over to $\displaystyle \mathbb{N}$ by f from M.
Thank you. However, I am still a little confused by the order carried over to N by f from M. The set M = {0, 1/2, 2/3, ...} But what kind of function should I use? The idea of (nonstandard) ordering of N confuses me, as I have never encounterd it before.
Thanks again.
It's natural to put 1 - 1/n from M into correspondence with n in ℕ, i.e., f(1 - 1/n) = n. The remaining elements are 1 from M and 0 from ℕ, so we pair them up as well, i.e., f(1) = 0. What order ≺ is induced on ℕ? By definition, f(1 - 1/n) ≺ f(1 - 1/m) iff 1 - 1/n < 1 - 1/m, i.e., n ≺ m iff n < m. However, n ≺ 0 for n = 1, 2, ...
Edit: Corrected n ≺ n to n ≺ m.