\b[Yes.] The induced order is strict and linear and remains dense. For $0<x<1$, the points $x/2$ and $(x+1)/2$ lie in the domain and are strictly below and above $x$. Hence the open interval is a <dense linear order without endpoints>.

Solution

Past exam of the logic and set theory course of the University of Oxford / 2010 / A / 3 / d / ii / δ / Solution

 Solution (source code)