Book i am following is Munkres.Suppose $X$ is well ordered set. Since every set contains smallest element , suppose i take any set of the form , say $(a,b)$ then consider set $(b,c)$ where $c>b$ .Since every set contains smallest element , let the smallest element in $(b,c)$ be $y$ .Hence $y$ is immediate successor of $b$ .Thus using this argument , every element in $X$ has immediate successor and predecessor .Thus if i take any set of the form say $(a,b)$ then it is closed automatically (by definition of order topology). Suppose i take any set of the form $[a,b)$ then it is also open since $[a,b) = (d,b)$ where $d$ is immediate predecessor of $a$ .
Thus by above argument we can say all intervals (open,half closed,closed) are open and closed in well ordered set .
Obviously i didn't proved for any arbitrary set and thus i am asking this question .
Edit : One more observation , every point set is open in $X$ .Suppose $x ∈ X$, then let $y$ be immediate predecessor of x , take interval $(y,x]$ , hence it is open by above argument .Is this observation correct ?
Best Answer
It does not follow that every element has a unique predecessor. For example, consider the ordinal $\omega + 1 = \omega \cup \{\omega\}$. Then $\omega$ itself does not have a predecessor. In particular, the set $\{\omega\} \subset \omega + 1$ is not open.