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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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 Math, Interior, Open set, Closure, General topology, Counterexamples in Topology

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