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Unformatted text preview: Solutions for HWK 5 INTRODUCTION Let B X . The set B B is called the closure of the set B . Notations: B or cl B . In what follows B denotes the interior of the set B . Recall that B def = { y : V y we have: V y B 6 = and V y B c 6 = } and that X = B B ( B c ) (1) (where the three sets are pairwise disjoined and some may be empty). Problem A Prove that x B iff for each neighborhood V x of x we have: V x B 6 = . (the condition is necessary). Let x B . It is clear that if x B then for each neighborhood V x of x we have: V x B 6 = . The same holds if x B , by the definition of the boundary. (the condition is sufficient). Assume for each neighborhood V x of x we have: V x B 6 = but x ( B c ) . This is impossible since x ( B c ) implies: V x B c so V x B = . Hence x / ( B c ) . By (1), this implies that x B B B B = B ....
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