MATH 140A Lecture 6: Note Oct 8, 2018
Let Sa Rbe aset so
M3So Max 5if socSand scSo Uses
upperlowerbound
sMis an upperbond of Sif SIM Uses
supAnf isupSsmallest upperband of S
inf Sgreatestlower band of S
Max1mm simple unique
may not exist
upper
flowerbands simple exist
notuniqueweak
intsup iexistuniquetight
not simple
The Completeness Axiom of IR
Hnonempty set ScRthat is bounded above has the
least upper bound iesupSECR exists
Def LR is defined as the smallestextension of IQ that
satisfies the completeness axiom
Remarkse aDef is non constructiveThese are construction of IR
using notion of completion
bcompleteness Axiom applied to 5sSES
yields the existence of Inf Se Raswell
Cor
4.5
study Vnonempty set Sc IR that is bounded below
PROOF hasthe greatest lowerbound i.e if SER exists
Equivalenttechmcaldefofsup.in M
oMsupSif iI
