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Homework 5 Solutions Katy Craig 2020 We proceed by induction Base case n 0 Co Eo D is the union of 20 1 disjoint closed interval of length 64 Inductive step Assume the result holds for Cn i By definition Cn is formed from Cn i by removing the open middle third of each interval Thus the number of intervals doubles and the length of each interval is a third of what it used to be Also all of the intervals remain closed and disjoint Thus Cn is the union of 2 2n 1 2 n disjoint closed intervals each of length E Est i CIM First note that part ensures Cnt BIR Hence C E BR E ell zoo

D union disjoint length - UC Santa Barbara

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Homework 5 Solutions

Katy Craig 2020

Weproceed by induction

Base case n 0 Co Eo D is the union of20 1 disjoint closed interval of length64

Inductive step Assume the result holds

for Cn i By definition Cn is formedfrom Cn i by removing the open middlethird of each interval Thus thenumber of intervals doubles and the

length of each interval is a third ofwhat it used to be Also all of theintervals remain closed and disjoint ThusCn is the union of 2 2n 1 2 n

disjoint closed intervals each of lengthE Esti CIM

First note that part ensuresCnt BIR Hence C E BR E ell zoo

IR 1 X

Also bydefinition CZ Cz Z and

2K Nco D L Thus by continuityfrom above

NC HAICnl nFo Hcn

line.AE HsT tinI.EsY O

First we show f is nondecreasing The problemstats that f is nonde

creasingon G DIC

Thus for all to t c Co DX with tottiIto Ef Iti

d

Fix x ye Co I withxcy.caseix yE0 Dlc ByLtd with tox try

fhdeflyCasey XECO DX and yEC Since XELo BK Kyf Ix E

ypgfit te LO DIC fly

Case3 X EC and yet DK If x O thensince ft 0 for all te fo DIC fade ftpSuppose x 0 By G with tryfor all to c o DIC to y

f Ito E fgSince to ex implies tody

f txt f It tea Dlc Efly

Case 4 exec and yEC

CLAIM There exists toEG Dl C sit XL todyPf of CLAIM Assume for the sake of contradictionthat for all to CCx

yto c C Then C

gontainsan open interval of le.ngth y x o

thus XD Vx yD y x 0 whichcontradicts

By the CLAIM and Cases 2 and 3f x Effto Ef

gNext we show f is continuous

Since f is nondecreasing Jiffy and gTx Kyexist and f is discontinuous at a

point Xo iff ligFxofly C yifxofly If xo O

we use the convention xoflg7 fCo and similarlyif Xo L we take ligIxofly f 1 I

Suppose f is discontinuous at Xo ECO D Then

finkoflyl glifxofly 4 f CEO D

However every open intervalin LEED

contains a point of the form In for some

KEIN n ft 2n D Thus I KEIN n ft 2n Dand X E fo DX sit

fix2k

E finkoflyl glifxofly

This is a contradiction Thus f is

continuous at Xo for all Xo E G D

Since Hx is non decreasing gCx fHtxis strictly increasing hence injective

Since ffx is continuous gtx fCxHx is

continuous By the intermediate value theorem

y

it attains all values in the interval

glo gaD 0,23 so it is surjective

To show g is continuous fix UE Lo B openIt suffices to show g D YU gCU is

open Since K Eo Dl U is compactandg

is continuousgk is compact

Thus the following set is open

0,231 gCK GEO DIK gluThis gives the result

ijective

First note that gktgnhcnlnfgknscn.gsa countable union ofKy

ftp.iiitnii gaingbijective

Since g is continuous gckin is compactThus ga E BR

Now by countable additivity2 260,231 160,23Igcc Nyce

i g i gA Thus it suffices to show 160,2 lg4D 1

To do this note that 8,231gcctgco.DKand recall that co Dic Ii whereIi are disjoint open intervals andf is constant on each interval i e fCIit Ci

Now

Neidigd Algae Dich Ngl tillH go.IE EFffitgiIIiD EHffIDtIil

translation invariance

EE Nci t Ii NII HE Iit No BKcountable additivity

By part b and countable additivity1 NEO D Ko Dlc tx c HEO Blato

Thus No 2 Igcc 160,1314 1 B CHthis gives

the result Y

By HW4 Q5 there exists kid s t

X Amc D 216 d Note ctd

It suffices to show E An Cgdsatisfies that E E contains an open interval

First note that Cc d Cc d containsthe interval l l l where l d c Wewill show f E E EE E This is equivalentto showing that t X Etf E there existsee ez EE s t ee ez X S e Eez TXIn other words it's equivalent to show thatE h E tx 0 for all xefE E

Assume for the sake of contradiction thatthere exists x El E E sit Eh Etx pSince EEE d Et x E c E ITE

Then bymonotonicityand countable additivityZe Nk E dt ET Z NEUE txNE t XEtx 2 XIE Il

4i

This is a contradiction

Recall from HW4 Q6 land as shown inclass there exists AEEO B with the

property that for all ER there is

exactly one YEA sit x ye Q

Also as shown in class A Heth To see this

suppose A Cethos and recall that we showed1 Hits

3 iguana 8 fee HE I

G implies 7CA 0 which contradicts GH

Forany a

beA a b is either 0 or irrationalThus A A does not contain an intervalso the result of part doesn't hold

It suffices to show h Ha to is open

for all a ER By definitionhe Ca t D gf ka tog which is a

countable union of open sets bythelower semicontinuityof F hence open

For any xot IR I Exo is lowersemi continuous