Question

1. Let (X, d) be a metric space, and U, V, W CX subsets of X. (a) (i) Define what it means for U to be open. (ii) Define what it means for V to be closed. (iii) Define what it means for W to be compact. (b) Prove that in a metric space a compact subset is closed.

Answer 1

1. Let (x,d) be a metric space and U, V, H C X subsets of x. each that said to open if for NEU there exists nyo such B (2) cu where Blag) open ball centred at padius 9 is with be V cx is it's complement said to closed if ve is open in x. is said to com pact if admits a every open Lover finite sub loven. (6) Let K be a lompact subset metric space (Xod). we Knon metric space a and Xos EX that every is Hausdroff "IR if xf y ex, then find (>0) such that we can where a is the B (2,8) n Bly, nult set. that k's Closed To open. prove prove that Let HEKE Then for each yek, using the

Hausanoff property of X, we can find Py such B (X, P4) n Bly, Ry) = 4. The collection { B(1, R): 4Ex} in <Bly, py): yfk} { B (40-50i),Jeison? { Bly, py): y6x}. of k. obviously open lovers Sine k is compact, so has a finite sublovey ret be a finite sub loven we is open. NON knon that every open ball is open set and intersection of a finite number open set Ba, 3;) is open set containing x. we claim that BOK 14 possible let at Brk. Then & EB and a Ek. Since KC Û B ( Yi, mi) So there ie such that XE B(Ys, 8;). = 0 exists some 4. [Mtk] Be B(aspi), for i= 1921-100 from we which since get XE B (2,8%) n B (Y5, 8;) [by using contradiction, REK' and Hausdroff property should but fr y Ek x & y so & hold for violates the Hausdorff property BOKE iR Bcke and 2 € ke such that open set NEBCKC. ont arbitrary Since

element of ke so ke is open Hence k is closed In a metric space a compact subset is closed.