Let \(X\) and \(Y\) be ordered sets in the order topology. Show that if \(f: X \to Y\) is order preserving and surjective, then \(f\) is a homeomorphism.
Let \(X = Y = \mathbb{R}_+\). Given a positive integer \(n\), show that the function \(f(x) = x^n\) is order preserving and surjective. Conclude that its inverse, the \(n\)th root function, is continuous.
Let \(X\) be the subspace \((-\infty, -1) \cup [0, \infty)\) of \(\mathbb{R}\). Show that the function \(f: X \to \mathbb{R}\) defined by setting \(f(x) = x + 1\) if \(x < -1\), and \(f(x) = x\) if \(x \geq 0\), is order preserving and surjective. Is \(f\) a homeomorphism? Compare with (a).
Solution
An order-preserving map in the strict sense satisfies \[ x<x'\implies f(x)<f(x'), \] so it is injective. Together with surjectivity, \(f\) is bijective and its inverse is also order preserving. Order isomorphisms carry open intervals and open rays to open intervals and open rays. Hence both \(f\) and \(f^{-1}\) are continuous in the order topologies, so \(f\) is a homeomorphism.
On \(\mathbb R_+=(0,\infty)\), the map \[ f(x)=x^n \] is strictly increasing and maps onto \((0,\infty)\). By part (a) it is a homeomorphism. Therefore its inverse \[ x\mapsto x^{1/n} \] is continuous.
The displayed map \[ f(x)=\begin{cases}x+1,&x<-1,\\x,&x\ge0\end{cases} \] is strictly order preserving and maps \((-\infty,-1)\) onto \((-\infty,0)\) and \([0,\infty)\) onto \([0,\infty)\), so it is surjective. But it is not a homeomorphism for the given subspace topology on \(X\). Indeed \([0,\varepsilon)\) is open in \(X\) for every \(\varepsilon>0\), while its image \([0,\varepsilon)\) is not open in \(\mathbb R\). Thus \(f^{-1}\) is not continuous.
There is no contradiction with (a): the topology on \(X\) in part (c) is the subspace topology inherited from \(\mathbb R\), not the order topology of the ordered set \(X\).