Talk:Equivalence of Definitions of Real Interval

Doesn't the "sufficient condition"-part of the proof only show that, given any $z\in\mathbb R$, there only exist $x$ and $y$ such that $S$ is an interval? --Matte (talk) 18:21, 27 August 2014 (UTC)


 * Now you mention it, it does look a little inadequate. --prime mover (talk) 18:25, 27 August 2014 (UTC)


 * Looking at ithe proof now, I think I misunderstood the assumption. Because it is assumed that $\forall x,y \in S:\forall z\in \mathbb{R}:x<z<y\implies z \in S$, it is sufficient to show that there exist $x,y$ in $S$ that satisfies the assumption for a given $z$ to be in $S$. --Matte (talk) 22:25, 31 October 2014 (UTC)