Question

plz use the definition solve the question

Definition 1. Given a set A CR, an elementu ER is an interior point of A if there exists an e > 0 such that (x - 5,3 +E) CA. The interior of A is the set Aº consisting of all interior points of A. A set A is called open if A= A'. Definition 2. Given a set A CR, an element X ER is a limit point of A if for every e > 0, there is some y EA c with y E (C-, +). The set of limit points of A is denoted A'. The closure of A is the set A := AUA'. A set A is called closed if A= A.

Answer 1

(a) Let & be any element of A. Therm PEA = A is a nbd.of 'p' there. enists EXO such that (p-,pre)CA Let q be any point of LP-E, P+E). Then (P-E, pte) is a nbd. of 'q' ( Every open set is a nbd- of each ? via point) Now, (p-6, pre) is anbd. of q and A LP-6,b+) => A anbd.9: Sunt į is any point of (P-E, P+E) Hence 'A' is nibd- of each part of (P-6,p+E) and so every point of . (p-6, pte) i an interis point of A. .. (P-E,P+E) CÅ (ia) PE(P-E, PIE) CAM As a nbd.fb. And p is any point of A, SO AO is a nbd. Of cach qit porümis. hence by the definition of A is an open set. (6) het A beany set. - Then the closure Ā AUA (where Al the derived set of A) = A ů a super set of A. Now we have to prone A i losed set T Ā closed s IR- à n open CIR- à is the compliment of à in IR) hat p ER- Ā tham Þ¢ Ā hance there existe €>o such that (P-, b+c) contains no points of A. Let'ą be any point of LP-E, pte) Since of CP-E, ptt) is a nbl. of a Containing no points of As it says that q¢Ã Sunce 2 is an arbitrary point of lp-E, P+E), we see that no point of (P-t, pte) can belong to Ā. Lp-tipte) CR- à =IR- sanbol. 0 But p' is any point of IR- à and so iR-A is mbo- of each of its limite.. Hence IR- ý an openset s à in closed.