Lecture 3

KatyCraig 2021

Recall

Deftfield A set F is a field if it has two

operations addition and multiplicationwhich satisfy the following properties forall elements a b CEFAI at btc at b t C associativityAZ at b b ta commutativityIA3 there exists an element in F

called 0 sit for all aEF at a a identityAU for each a c F there exists an

element called a C Fast at a 0 inverse

M1 alba ab c associativityM2 ab ba commutativityMS F an element in f called 1

St 1 10 and Va EF a f a identityM 4 for each AEF at 0 there exists an

element called ta EF s t a ta L inverse

distributiveDL allot c abt ac Iaf

Def ordered field A field F is an ordered

field if it has an ordering relation E

so that for all a b c EF

Ot either a E b or be a totality1021 if a c b and bea then a bantisymmetric03 if a Eb and bec then a E c transitivity1041 if a

c b then ate Ebtc addition05 if AE b and C 20 then a CE b e multiplication

On an ordered field wemay define

notions

of maximum Iminimum bounded above below

and supremumlingimum

Deff maximum minimum For SEF Max s is the

largest element of S and mines is the smallest

element of S M is an upperbound

gsDefboundedabove below For.SEif F Me F s t H se S s E m then S is boldaboveif F me F s t V s E S s z m then S is bold below

m is a lower bound ofsIf S is bold above and below then it is bold

Def supremumlinfimums For SEF the supremumofSs the least upper bound of S and the infinumgsis the

qfqatest.to er bEu.nd of S

i e if G so so is a lower bound 5 that is so Es Vs ES

so is 2 any other lowerbound

Ex If 5 Ca b we will soon prove that

infls min s a

sup s b though the set has no maximum

Re makC gumming

Wheneverthey exist max G ES and mints ESOnthe other hand supls and inf G don't

necessarily belong to S EF for Fan offended

Any finite set 5 se Sz Ss Sid always hasa minimum and maximum

aExerciseGi ven an ordered field F and S E F

If Max S exists then sup Lst mailsIf min S exists then infls min s

Because of this the supremumlingimumis a generalization of the notion ofmaximum minimum

Def Creal numbers the set of realnumboisis the ordered field containing Q withtheproperty that

every nonempty subsetSER

that is bold above has a supremum

The Least Upper BoundProperty ofreal Numbers

We'll study two major theorems for RArc Eropotty and Qisdense.in R

MAJOR RESULT 1

Them Archimedean Property If a b E IR

satisfy a 0 and boo then there exists NEIN

so that na bTspoon bathtub

Rem ak Even if a isreally small

andb is huge some integer multiple of a isbigger than b

Given enough time one can emptya

large bathtub with a small spoon

we will prove by contradiction

Scratchwork

D ta b O F n HN s t na b

P Fa b O s t V nel N na E b

Pf Assume for the sake of contradictionthat F a be IR with a O b O S t

for all new na Ebµa Za Ba 4A

Define 5 na NEIN so b is an upperbound for S Since S is a nonempty subset

of R that is bounded above by olefin ofRS has a supremum Define so sup S

Since a 0 we have so a Soc SotaSince so sup

s there exists no C INS t So a no a so thot Da

Since thot1 a ES this contradicts the

fact that so is an upper bound of S O

As a consequence ofthe Archimedean

Property we have a few useful lemmas

Lemme For anya EIR there exists NEIN s t asn

Pf If at 0 then the result holds for relIf a58th'T then since 15 9 by AP

0there exists me IN s t I n a

Lemmaj For any a.be Racts there exists NEIN

so that at the b

ftp.attnI 1 3Mental image a b

spoonbathtub

Pf Let y b a O and I 0 By APthere exists me IN Stmy

IyI

b a In at In b O

Jemima If x.ge R satisfy Kxy then

FMEZso that y mix

myMental image

C I I 3

Y there shouldbe an integersomewhere in here

Pf By the first Iemma there exists me INs t n

y Define 5 jez y jenThen S is nonempty and finite so

m min S exists By defn of m MEZ

yarn and m leg Thereforeyam Etty X O

Now we will apply the previous lemmasto proveMAJOR THEOREM 21

The CQ is dense in IR If a bHR with acb

there exists r C Q satisfying a r b

r C Q

Mental imageL l l l 7

a EIR BEIR

This is similar to the result we proved on thefirst day that between any two rationalmembers there is a ratiberT

PffBy the lemma F nel NS.t.attncbnatl Cbn Isbn an By theother lemma there exists in E Z so

that ane m bn a In Lb O

We now have all the tools we need to

rigorously prove our previous claimsabout the minimum Imaximumlingimumlsupremum of subsets of R For example

Prep For a b E IR a b the set 5 Ca bdoes not have a maximum and sup s b

2 1,93F st we show that S does not have a

maximum Assume for the sake ofcontradiction that max G MoSince Moes F Mo b

By densityofIQ

in IR F r E Q s t Moc r e b so r ESThis contradicts that Mo was the

largest element in S

Now we show supG b By defin of Sb is an upper bound SuppsoomsEa Mo isanother upper bound of S If Mo Lbthen by densityof IQ in IR F r E Q s t

Mo L r L b s o r E S which is a contradictionThus Moz b so b is the least upper bound O

Goingforward we will use to and to

simplify our notation for Suprema and infima

Ex 1a ta x EIR ac x x EIR a xd to

Def Unbounded above below Suppose SECR isnonemp

If S is not bounded above write supG too

If S is not bounded below write infa

Remark Given anonempty

SEIR

Bydefn of supremum and R

S has a supremum S is bounded above

supG ER

SimilarlyS doesn't have a supremum S is not bonnadggle

supGto

Using this notation even though not

every sethas a supremum for any

nonempty SEIR sup G has meaning