Question Number 192336 by Mastermind last updated on 14/May/23
Answered by Rajpurohith last updated on 27/May/23
$${Since}\:\boldsymbol{{S}}\:{is}\:{bounded}\:,{inf}\left(\boldsymbol{{S}}\right)\:{and}\:{sup}\left(\boldsymbol{{S}}\right)\:{exist}\:{and}\:\lambda\in\mathbb{R}. \\ $$$$\left({a}\right)\forall{s}\in\boldsymbol{{S}}\:,\:{inf}\left({S}\right)\leqslant{s} \\ $$$$\Rightarrow\:\:\forall{s}\in\boldsymbol{{S}}\:,\:{inf}\left({S}\right)+\lambda\leqslant{s}+\lambda \\ $$$$\Rightarrow{inf}\left(\boldsymbol{{S}}\right)+\lambda\:{is}\:{a}\:{lower}\:{bound}\:{of}\:\boldsymbol{{S}}+\lambda. \\ $$$${if}\:\:{inf}\left(\boldsymbol{{S}}\right)+\lambda<{t}\:\left({be}\:{a}\:{lower}\:{bound}\:{of}\:\boldsymbol{{S}}+\lambda\right) \\ $$$$\Rightarrow{inf}\left(\boldsymbol{{S}}\right)<{t}−\lambda\:\left({which}\:{is}\:{a}\:{lower}\:{bound}\:{of}\:{S}\right) \\ $$$${which}\:{is}\:{a}\:{contradiction}\:{since}\:{no}\:{such}\:{lower} \\ $$$${bound}\:{which}\:{exceeds}\:{inf}\left(\boldsymbol{{S}}\right),\:{can}\:{exist}. \\ $$$${Thus}\:,{inf}\left(\boldsymbol{{S}}+\lambda\right)={inf}\left(\boldsymbol{{S}}\right)+\lambda \\ $$$$ \\ $$$$\left({b}\right){let}\:\lambda>\mathrm{0}\:\Rightarrow\lambda{s}\leqslant\lambda{sup}\left(\boldsymbol{{S}}\right)\:,\:\forall{s}\in\boldsymbol{{S}} \\ $$$$\Rightarrow\lambda{sup}\left(\boldsymbol{{S}}\right)\:{is}\:{an}\:{upper}\:{bound}\:{of}\:\boldsymbol{{S}}_{\lambda} . \\ $$$${let}\:{u}\:{be}\:{any}\:{upper}\:{bound}\:{of}\:\boldsymbol{{S}}_{\lambda\:} {such}\:{that} \\ $$$${u}<\lambda{sup}\left(\boldsymbol{{S}}\right)\:\Rightarrow\frac{{u}}{\lambda}<{sup}\left(\boldsymbol{{S}}\right)…\left(\mathrm{1}\right) \\ $$$${By}\:{our}\:{assumption}\:{that}\:{u}\:{is}\:{an}\:{upper}\:{bounf}\:{for}\:\boldsymbol{{S}}_{\lambda} , \\ $$$$\lambda{s}\leqslant{u}\:{which}\:{shows}\:\frac{{u}}{\lambda}\:{is}\:{an}\:{upper}\:{bound}\:{of}\:{S}\:{which}\:{does}\:{not}\:{exceed}\:{its}\:{supremum}\:\left({by}\:\mathrm{1}\right) \\ $$$${Hence}\:\:{a}\:{contradiction}! \\ $$$$\left({c}\right){Hope}\:{you}\:{can}\:{try}\:{further}. \\ $$$$ \\ $$
Commented by Mastermind last updated on 04/Jun/23
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$