Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 192336 by Mastermind last updated on 14/May/23

Answered by Rajpurohith last updated on 27/May/23

$$Since\mathit{S}isbounded,inf\left(\mathit{S}\right)andsup\left(\mathit{S}\right)existand\lambda \in \mathbb{R}.$$$$\left(a\right)\mathrm{\forall }s\in \mathit{S},inf\left(S\right)⩽s$$$$\Rightarrow \mathrm{\forall }s\in \mathit{S},inf\left(S\right)+\lambda ⩽s+\lambda$$$$\Rightarrow inf\left(\mathit{S}\right)+\lambda isalowerboundof\mathit{S}+\lambda .$$$$ifinf\left(\mathit{S}\right)+\lambda <t(bealowerboundof\mathit{S}+\lambda )$$$$\Rightarrow inf\left(\mathit{S}\right)<t-\lambda \left(whichisalowerboundofS\right)$$$$whichisacontradictionsincenosuchlower$$$$boundwhichexceedsinf\left(\mathit{S}\right),canexist.$$$$Thus,inf(\mathit{S}+\lambda )=inf\left(\mathit{S}\right)+\lambda$$$$$$$$\left(b\right)let\lambda >0\Rightarrow \lambda s⩽\lambda sup\left(\mathit{S}\right),\mathrm{\forall }s\in \mathit{S}$$$$\Rightarrow \lambda sup\left(\mathit{S}\right)isanupperboundof{\mathit{S}}_{\lambda }.$$$$letubeanyupperboundof{\mathit{S}}_{\lambda }suchthat$$$$u<\lambda sup\left(\mathit{S}\right)\Rightarrow \frac{u}{\lambda }<sup\left(\mathit{S}\right)...\left(1\right)$$$$Byourassumptionthatuisanupperbounffor{\mathit{S}}_{\lambda },$$$$\lambda s⩽uwhichshows\frac{u}{\lambda }isanupperboundofSwhichdoesnotexceeditssupremum\left(by1\right)$$$$Henceacontradiction!$$$$\left(c\right)Hopeyoucantryfurther.$$$$$$

Commented by Mastermind last updated on 04/Jun/23

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com