Let \(X\) be an ordered set in the order topology. Show that \(\overline{(a, b)} \subset [a, b]\). Under what conditions does equality hold?
Solution
The set \([a,b]\) is closed in the order topology: its complement is the union of the open rays \[ (-\infty,a)\cup(b,\infty). \] Since \((a,b)\subseteq[a,b]\), minimality of closure gives \[ \overline{(a,b)}\subseteq[a,b]. \] All points of \((a,b)\) are already in the closure, so equality is equivalent to having both endpoints \(a,b\) in the closure.
The point \(a\) lies in \(\overline{(a,b)}\) exactly when \(a\) has no immediate successor in \(X\) (equivalently, every order-neighborhood of \(a\) contains a point strictly between \(a\) and \(b\)). If \(a\) has an immediate successor \(s\), then an order-neighborhood ending at \(s\) contains \(a\) but no point of \((a,b)\). Conversely, if there is no immediate successor, every neighborhood of \(a\) contains some point \(x\) with \[ a<x<b. \] Similarly, \(b\in\overline{(a,b)}\) exactly when \(b\) has no immediate predecessor.
Hence \[ \boxed{\overline{(a,b)}=[a,b]} \] if and only if \(a\) has no immediate successor and \(b\) has no immediate predecessor.