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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?

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I'm not sure it does; take $\mathbb{Q}$ and give it the cofinite topology (so a set $A\subseteq\mathbb{Q}$ is open if and only if $A=\emptyset$ or $\mathbb{Q}-A$ is finite). Every interval of rationals is connected in this topology, because any nonempty open set intersects any infinite subset, but $\mathbb{Q}$ certainly does not have the least upper bound property. –  Arturo Magidin Dec 15 '10 at 4:50
    
@Marcus: But then we don't suppose that every interval is connected; every interval is connected. Are you trying to see if one can prove the least upper bound axiom purely from the topological properties of $\mathbb{R}$, perhaps? –  Arturo Magidin Dec 15 '10 at 5:00
    
@Arturo: in your example, the interval is the set that is bounded above, so doesn't it has a least upper bound? I'm confused, you seem to take $\mathbb{Q}$ as as counterexample but $\mathbb{Q}$ is not bounded above. Am I missing something? –  Marcus Dec 15 '10 at 5:10
    
@Marcus: I was giving you an example of an topological set that has a partial order (not related to the topology) in which every interval (in the usual sense) is connected but which does not have the least upper bound property. Take any interval in $\mathbb{Q}$, then it is connected in the cofinite topology. But we know that $\mathbb{Q}$ does not have the least upper bound property relative to the usual order. As Pete Clark now answered (I was in the middle of a reply which his makes silly), if the topology is connected to the order, then the answer is yes. Your use of "suppose" confused me. –  Arturo Magidin Dec 15 '10 at 5:17

2 Answers 2

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.

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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.

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