Question

4. Let A = [0,1) CR, where R is endowed with its usual metric. (a) What is the interior of A? Prove your answer. (b) What is the closure of A? Prove your answer.

Answer 1

Gillen A=[o, 1 CR Where (R is endowed with whal metric ie, netric defined as the usual distance la Llu,y) = (x-4) NOTE metric Interior boint is same as Close of set to find interior in l endowed with usual point in R and closure of set Interior point s Let & ER and sc R 2 io sitob an interior point of s if so sit (2-8, 2+8) cs. lie & is an interior point of s iff sis a nobl of a ZES. A = [0, 1) Let & E (011) clearly, o ddato 1 for cay E (or) (2-5, 2+s) c [112) ( here rocan be choose as small as we can ) I Nowe take &=0 ofd Co-sotd) & 1011) o is not an 0-8 to a interior point,

Now take dat Now take a (1-6, 1+8) ¢ [0, 1) 1 is not an interior boint. let & E (R- [0, 1) Clearly it can be seen that (2-8, 2+8) ¢ [a 1) Hence, & ER-[0, 1) is not an interior point. Å = Interior of A 6 Closure of set = sus wel where s'= Set of Accmulation point of Sant Accmulation Point Let & E R and SSR & is s. tab accmulation point of siff (2nd, 2tdns is infinite. ) A= [0, 1) & E (011) Clearly from figure (2-8, 2+d) nA = infinite. • LY LEA Now L=0 (6-5,0+8) nA = o E A and oth a 10, ots) = infinite

Now take 2=1 (1-d, 1+8) nA = (-4,1) = infinite 1 E A Now take & E (R-[0, 1] (2-8,2+8) n A = empty & infinite && A 22613 = Hence, A= (0, 1) u {o} U {1} Ā - AUA = [0, 1] NOTE Ã stands for closure of A sotib stands for said to be abd stands for neighbourhood