Does the fact that every interval in $\mathbb{R}$ is connected implies that $\mathbb{R}$ is order-complete?

Question

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?

Answer 1

The answer is yes, in the following sense:

Theorem: For a linear order on a set $X$, the following are equivalent:
(i) $X$ is connected in the order topology.
(ii) The ordering is dense and $X$ is Dedekind complete -- every subset which is bounded above admits a least upper bound.

See Theorem 13 of these notes for a proof which also explores the connection to induction in totally ordered sets.

Answer 2

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.