Wulf ExerciceFor the représentation there are two ideas

first when there are several conditions

the and also denoted Ï correspondsto intersection and the er ton union

it allows to decompose the set intosoinplier onesSecond when you have a set with

inequality conditions libre fluideTRYn'taie 13 you begin to studythe equality case in general it will

given you the boundary of the firstset and you find the set by takinga particular point the boundary being

in the set a not depending ifthe inequality is strict or not

1 A nn c R sitter A laideur n o

n Cnn C KE y o

For A1 you consider ln g c W Ntfswhich gives a unit aide centered at

O D and which is the boundary of AI

A1 is then the disc

Æ

Æ

EAAAz As AI

A is bounded because la µ c AIl lush Il E 1

Ae Az and Az are closed as preuûages ofclosed intervals by continuous functions

fa example At f YEE whereclosed

flank n 4 1 is continuous as a polynomial

idem for Az and AzThen A is closed as intersection of closedsets

A is compact because closed and bounded

in R I finite dimensional space

forthe exam you should give soûilararguments for the next questions we justgive the solution

y R

y n

B not bounded because n o c B tn 1

and k ch o Il to

B not closed because cut o c B tu 1

tout LE d go et Bhosts

B not open because 1 de B but there is

not e o l B e c B since

Hero 11 f e Bla I e but not in B

Burt compact because in R2 a subset

is compact d is bounded and closed

3

C not bounded since n e c c tn o

but Knew o

C closed C f Co toc where

closed

floyd y_et is continuous because

x et continuous and ln ul y continuous

Ü for example Co 1 e c but

there is no Eso I Bligh e CC

because Hcg 0,1 E e BUN c

and is not in c

Cnrt npact l sane argument than B

D not bounded lanice nif c D

and It ln Il a Thushrtempnot

D not open 11,1 c D tut there is

not e ol Blah e CD since

11,1 E D te o butin Blait

D closed if cnn.int ED converges to

nm yn 1g nn o

xn converges to x o but n o becauseotherwise yn would converge to

to l and not toy Thus

yn1 but abryn y

to K hosts

Thus y In no i e qu c D

Thus D closed

Exercises A

ÎÎÈÜ

fËÏÏÏ

because po c B C yo tB but

4dtf qu et B

convex if

and c Cop then y n

y ni

ntll H.nl s n a Huit because

n n convex function since

Cnil 270 Un c R

s y a a y y

Thus Amy LI Hoi De C

Exercicef1 lut tn on Ar 70 toc

fr not bounded above on A1 because

fr f to and the A1hosts

Thus Sup f141 tone 11

fr bounded below on Ar since

f1 n 70 Vu c A1 Thus

Inf f4 exists à Rne A1

Also he AI Vu 1 and

film 0 Which is a muinant

of fon At From the sequential

characterization of vif this proves

Inf film o

neAlas fzln not bounded above suice

lin feld to Thus sup fin tanato ne AL

fa x n 1 _1 canonical form

Thus fut z 1 tuer i e

1 is a minorant off ou Az

But fr111 1 Thus

Inf fin Min fr14 1ne AL REALstrictly

lut is P on Cpac

Tf Kui II I

l lena l Il

Thus