Let < be a lexicographic order on $^{\omega}2$ or in other words given distinct functions $f,g$ from $\omega$ to 2, let $f<g$ if and only if $f(n)=0$ and $g(n)=1$, where $n$ is the lease $m<\in\omega$ such that $f(m)\neq g(m)$. How can we see ($^{\omega}2$,<) is not well-order.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ If we throw away the functions that end in an infinite number of ones, then the remainder of your poset is isomorphic to $[0,1)$ with the usual order by treating the functions as binary expansions, so well-orderedness fails very badly indeed. $\endgroup$– Ben BarberApr 22, 2014 at 15:18
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let $(f_n)$ be the point that maps $\{0,\dots,n\}$ to $0$ and the rest to $1$. Then $f_{n+1} < f_n$ for all $n$, showing an infinite descending sequence.