TITLE: Does the fact that every interval in $\mathbb{R}$ is connected implies that $\mathbb{R}$ is order-complete?
QUESTION [6 upvotes]: Suppose that every open interval in $\mathbb{R}$ is a connected set. Does this implies the least upper bound axiom? (i.e every non-empty subset of $\mathbb{R}$ which is bounded above has a least upper bound) Is this true? in such case, how would you prove this?

REPLY [2 votes]: Suppose not.
Take two intervals which witness that fact - namely $(a,b)$ and $(c,d)$ such that $b<c$ however $(b,c)=\emptyset$. 
Choose $\varepsilon$ small enough and look at the interval $(b-\varepsilon , c+\varepsilon)$ which is nonempty. By our assumption this interval is also connected, there is someone between $b$ and $c$ otherwise you'd have a non-empty interval which is not connected. A contradiction.