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# Contains a point x s x 6 b since open balls in r are

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contains a point x S , x 6 = b . Since open balls in R are intervals, this is the same as proving: For every > 0 there is x S , | x - b | < , x 6 = b . So let > 0 be given. Then b - < b , so b - is not an upper bound of S ; there exists thus x S , b - < x b , in particular | b - x | < . Since b / S , we must have x 6 = b . 5. Let X be a metric space and let p X . Let C p = [ { C : C is a connected subset of X and p C } . In other words, C p is the union of all connected subsets of X containing p . Prove that C p is connected. Solution. Since { p } is connected, this is of course an immediate conse- quence of the second Proposition on page 60 of Rosenlicht. A direct proof is also quite easy, if one followed the proof of that proposition given in class: Assume C p is not connected; there exist then open subsets A, B of X such that C p A B , A C p 6 = ∅ 6 = B C p and A B C p = . Notice that p C p ; in fact, p ∈ { p } and { p } is connected, thus { p } ⊂ C p . Then p A or p B ; without loss of generality we can suppose p A . Assume now C is a connected subset of X such that p C . Then C C p A B ; C A B = since it is a subset of C p A B = . Moreover p C A so C A 6 = . The only way we can avoid contradicting the fact that C is connected is by concluding that C B = . But since C A B , this implies C A .
• Fall '09
• Schonbek
• Topology, Metric space, Topological space

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