Exercise 7, § 2
General Topology, IV
Bourbaki marks this one as harder than the rest.
Let be a non-empty linearly ordered set, endowed with the topology .
a) If is connected, show that it has the following two properties:
α) Every non-empty subset of which is bounded above has a least upper bound (if is non-empty and bounded above, and if is the set of upper bounds of , and the set of lower bounds of , show that and that and are closed).
β) The set is without gaps, in other words (Set Theory, Chapter III, § 1, Exercise 18) every open interval of is non-empty (use the argument of Theorem 4).
b) Show conversely that if has properties α) and β), then is connected. (Show first that every closed interval is connected: assuming that this interval is the disjoint union of two non-empty closed sets and , and that , consider the greatest lower bound of and hence obtain a contradiction).
c) Show that if is connected it is locally compact and locally connected, and that the only connected subsets of are the intervals (bounded or unbounded).
Solution
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