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?
Does the fact that every interval in $\mathbb{R}$ is connected implies that $\mathbb{R}$ is order-complete?
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real-analysis
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order-theory
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0I'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. – 2010-12-15
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0@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? – 2010-12-15
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0@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? – 2010-12-15
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0@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. – 2010-12-15