Question

Let A be a subset of R. If x € R we say x is a boundary point of A if for all € > 0, (0 – €,x +E) NA # and (x - €, x+) NĀ+). The boundary of A is a A. It is the set of all boundary points of A. The interior of A is int A = A (BA). The closure of A is cl A = AU (DA). Let B be a subset of R. Prove the following statements: (Hint: Do them in order and use results of previous questions in your solutions.) 1. The interior of B is the set of all interior points of B. 2. OB = B; int B, BB, and int B are disjoint; and R = (int B) U (AB) (int B). 3. int B = cl B and cl B = int B.

Answer 1

2 1 Gyven BSB Prove that interior of B is the set of all- interior points of B. ( Note! - Interior point ! Let Acir, ner is said a to be interior point of A if s>o such that I (a-s, ats) na 74 ie. Prove that Int (B) = { x Girl a is interior point of B} Let at Int (B) =) 2€ Al(SA) = n is not a boundary point By using negation of boundary point, there exigt eso sit (x-e, at en A= a is an interior point of B. 2 is an arbitrary elemento {xeir l ais point Int (B) c interior of B} - Now, we prove Int (B) ? {NER 1 X is an interior point of B} let a {REIRI a is an interior point? 3 X )'s an iferior point. By cefn of interior point Isso such that (a-s x+5) n B = 6 Negation of art above statement Vsoo sito (a-sa,.&+5) AB 0 xe Int IB).

2) prove that a B = 2 B Let acaB YEJO La (2-6, ate) O p = 0 and la-e,*+) = D (96, 9+E) * = and (-6,4+) 0 B = 0 XE OB 2B = 2B Now, we prove int B, a B and int B are disjoint. we prove it by contradiction let no intB na B o inte 7 ne int B and ne ab, and ae int B ne int B, by cloth of interior point 7870 St. (x-S, 2+5) 0 B =0 - @ NedB, then teno la-E, a + E) n B & 0 and (a-e, ate ) noto - this is true for all e choose E zł 9 then (2-5, 2+s 70 BE $, and (2-6, ate)n Be from CD & 67 , it is impossible. Hence our assumption was wrong > Int B na B=0

. a E Int B I go such that (-5', 2+s') 0 B = 0 -64* From + and on, it is impossi Our assumption was wrong - . O Boint B = 0 intB na B o intB=0 :. Inst B, aB and int B are disjoint. then clearly, IR = (int B) U2B) u (Int B) cinta ( B = IR-B)

3) Int B = cd B Let ne Int B = ae a B u int B Int B = 2 B U intBZ à e a B U in i s int B e B } ne cl 3 ( cd (B) = OB UB ) : Int B c d B Now we prove cl B C Int B Let at cel B at a BUB If RedB ar E Int B It ne B : cel(B) then ne Tnt B Int B Int B = :: Cd B

B (l(B) = int B let ne cel(B) = ae BulaB) Be & XE B n JB B = nEB O (BU Int B) : dB = BU Int B ? 7 nEBnB) u/BnInB) 2) a E Int B (from figure). cel (3) c int B let ne Int B 3 a ¢ B U2B) a cl(B) : Ca(B)= BUAB ? = ae cl(B)