Question

Please answer question 3.21 and 3.22. Thanks！

3.21 Prove the following facts: a) For fixed ECR, the function d(x, E) is continuous. b) If E and F are subsets of R, then d(E,F) = inf{dy, F): y EE}. c) If ACE and BCF, then d(E,F) <d(A,B). d) d(E,F) = d(E,F). 3.22 Prove the following facts: a) Suppose that F is a closed subset of R, K is a compact subset of R, and FOK = 0. Then d(F,K) > 0. . b) Show that part (a) is not true if it is assumed only that K is closed.

Answer 1

| 3.217 a) let Ecp. Then the function I FRIR defined by i s i - fx=d(a, E) . We shall prove that fro= d (1,5) - fisa continuous functions Proof let dot R be arbitrary. not R be arbitrary and to be that pre-assigned. Then it n it is enough to show is enough to sho at B(,)= |(1,7) – 4 (40,F)] <<. suppox XE B (2, 'f) and at E be arbitrary a Then d (1, 0) < 2 (2,8.) + d(*0,)< < td(90,€) => inf ſocora): ac < + inff 2(4,8) : GE} red(2, F) < d (10, E) +€. Intercharging x and to it we get d(20,5) < d (1,5) +ę. Thus 1d (1,F) – d(40, E)].<ęce. This f is continuous function. [Porover]

we Kwas that d (EF) = inf (d(my) ; ret,847) Now d (»;+) = inf (169,2): 26F? ic d (0,5) = ing as d (1,9): 9ef} allow see. Now tare inf d(045) = inf (d(1,9) : 9745,44F). ne y This d(E,F) = inf ( 2 (,F) : ate? = inf (d (y,F): YtE? [Proven e IF ACE, and B CF d(A,) = inf (4,1) : (6,6)€ Axol Then het to chosen arbitrarily Then I (ao, bo) E A7B The d (A,B) + c d (20,5-). Now d (ao, b.) < d (A, B) TE Since (a., b) € (AXB) < ExFi This d(E,F) sd (ao, bo) < d (A,B) tt → d (E,F) sd(A,B) tk. . 1 Since E is arbitrary thus. d(E,F) SJCA, 9)

7 It is clear that d (E,F) sd CE, F) Ed (EF) - Note that the infimum of hiyyeye set is smaller). We only show that d (FF) < d (F1) Let xfEYE F and let so be arbitrarey. " I afE and bff such that d(a, a) < & and d (y, b) < This d (a, ) < d (a,x) + day) + d (9) << + d(», y) ++ d (,) + → dient Hoot 04/+4 d (F/F) < d (a,b) < d (yy) +tips > $(1,7) » d (E,F) - f t net and ye F 9 d (EF)>d(E,F) - E. 3 d (E,F) sd (EF)+t → d (E,F) < d (TF). Since trois arbitrary. This d (EF) = 2 (E,F). [proved)

20) F is a closed subset of R and K is a compact subset of R and Enk=9. Then we shall show that d E, F) . Since F is closed and Enf=a. for each atk d (a,F) >0 [as Fis closed and a¢ F). Define f : R→IR by fax = d (^,F) V ER. Then f is continuous for each atk f(a)=d(a,F) >o. We choose for each atk a real number ra satisfying ocras fras. Then Sf (ra, a): at klis I an open cover of k [since f is continuar (rap) is open in IR for each ack). a kis compact , finitely many points a,a2, ant ko such that ke ů s ' (ra, a) .net 1 Nax = min (ra. : isisny. Then rap to and k cf Craxia) ie d (a,F) > Fay Nath and hence inf d(a,F) Sirax in d (K, F) 7 Nax7o. . [Proved R atk 3 If Kis closed and Fi closed with KnF = 9. I bet k= N and F= fant 1 Enton] - h t o and k and F o both are I Closed. set with knF=Q. but d(x,F) = 0.