Question

1. Draw the given sets. Find their interior, exterior and boundary and also sketch them. Moreover, study whether these are open, closed, bounded and/or compact. (0) A = (3.7]{0,1} E R. (ii) B = {(x,y) ERO SysIn... ISIS 2). (iii) C = {(x,y) E R' + ly <l},

Answer 1

Here we use a famous result to prove the compactness of given sets. Heine-Borel theorem: The compact subsets of $\mathbf{R}^n$ are precisely the closed and bounded sets of $\mathbf{R}^n$ .

I have in detail explained the open, closed, interior, exterior, boundary, bounded, compactness, and plotted corresponding graphs. Kindly go through.

HC A = [3, 4] u {0,13 € Ro b dubio Interior of A - (3, 1) i l i Proof:' Let tė (3:2) C let v= min { 1 x=3), lo ==1} Thers (x-7, **») c (3,7) c A But 01, 3,7 are not interior points of A. Since if we take any neighborhood of o, 1, 3, it there will be points that are not in A. Exterior of a set s is the set of all points that are meither in Sned in its boundary Here, boundary print means, mi is a boundan point of A iff every nghol of a contains points from both Å and A". Boundary of A = {0,1,3,7. Exderiori e of A = R-A = (-010) V(0,1) U(13) (7,0).

Interior of A: B ? 3 Exterior of Arts Bil 3 Bourndary w of A: I 3 Since 3 is heat an interior poist, and she seA, A is not open. (A is open off A- A°) Å is closed since A has all its boundary points. powro This he A is bounded below by - and above by 8 Hence A is boun ded. " By Hieni borel Theorem, closed and bounded subsets of Rh are compact. Herse cs And Compact T

(6) B = {Grydek | osya lux , !Exe2} ... Draw y= lux and y=o. Also a-1 and x=2. The region bourshed by these curved is B. NO yolux 8= 1 2 =2 tt B. P anel, Ima ge 0 1 2 - limenier of. B: { Gyder*1e3c2 , oxy< lnz} (2, 2s 2) (NOTE: line segment I to 2 is not illud included), CS Scanned with CamScanner

. Exterior Bi Note: linegegnet Ho 2 is not included Boundary B : ON <3, line) ={GyJcR" | rexse, groju {{wydeR* Log£ luz ,x=} u{@y?ER? ya lnxe, 15x52}. We have (1,0) é B but not an interior poista This B is not ofen. . since (boundary of Bc B B is relosed. Also Bc B ((0,0),5). So B is bounded. Since B is closed and boundled subset of Bhandabai. Hieneberel. Theorem, B is compact U porel Compact rem , un nScander

REA () . C= {eysera | 1x1+ly!<1}. How to draw. C C is bounded bena. 4 lines x+y=1, r=y=1, -x+y=1, -x-y=1. The bounded region withouth these lines will give me Ciao patients Cr (1:0) 61,0) poist of Ċ is an none Interior of C: We have any interior poinst of c... Interior of ci 1 Scanned with CamScanner

Exterior of C: sxterior of Since every Ntu = point on the lines x+y=1, a-y=1, are boundary poists, Exterior of C: { Bayer") Kry>13 ufanyer“) a yo'} {my)ER "/x+y=1} & {y) eR" |-x-y=1}, Buralary of C: * from = {BYDERT / Bal Hyl=1. Tot llo) Since (interior of C)= c, Cis ofers. (1,0) is a boundary point of c. But (1,0)&c. Hence c is not closed. Cis bounded since cc B (10,0), g). By Hienborel theorem, since C is not closed C is not compact. 160,-1). OV sel Scanned with canner, FS-

The hole or dotted line in the graph represents that the point does not belong to the corresponding set. The required regions are shaded with different (sometimes same) colour. Since the x-axis must always be drawn in a graph, it has been drawn even when it is not in the required set. There I have mentioned in brackets specifically. Kindly consider it.