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Question Number 192299 by Mastermind last updated on 14/May/23

Answered by witcher3 last updated on 14/May/23

R⊆S⇒x=Sup(R)≤sup(S)=y ∀r∈R r≤y,∀ε>0 ∃s∈S such y−ε<s≤y by definition ∃r∈R r≥s ⇒∀ε>0 y−ε≤r ε=(1/n)⇒∀n∈N y−(1/n)<r⇒s≥y⇒sup(S)≤Sup(R) ⇔sup(R)=Sup(S)

$$\mathrm{R}\subseteq\mathrm{S}\Rightarrow\mathrm{x}=\mathrm{Sup}\left(\mathrm{R}\right)\leqslant\mathrm{sup}\left(\mathrm{S}\right)=\mathrm{y} \\ $$$$\forall\mathrm{r}\in\mathrm{R}\:\:\mathrm{r}\leqslant\mathrm{y},\forall\epsilon>\mathrm{0}\:\exists\mathrm{s}\in\mathrm{S}\:\mathrm{such}\:\mathrm{y}−\epsilon<\mathrm{s}\leqslant\mathrm{y} \\ $$$$\mathrm{by}\:\mathrm{definition}\:\exists\mathrm{r}\in\mathrm{R}\:\mathrm{r}\geqslant\mathrm{s} \\ $$$$\Rightarrow\forall\epsilon>\mathrm{0}\:\mathrm{y}−\epsilon\leqslant\mathrm{r} \\ $$$$\epsilon=\frac{\mathrm{1}}{\mathrm{n}}\Rightarrow\forall\mathrm{n}\in\mathbb{N}\:\:\mathrm{y}−\frac{\mathrm{1}}{\mathrm{n}}<\mathrm{r}\Rightarrow\mathrm{s}\geqslant\mathrm{y}\Rightarrow\mathrm{sup}\left(\mathrm{S}\right)\leqslant\mathrm{Sup}\left(\mathrm{R}\right) \\ $$$$\Leftrightarrow\mathrm{sup}\left(\mathrm{R}\right)=\mathrm{Sup}\left(\mathrm{S}\right) \\ $$

Commented by York12 last updated on 14/May/23

sir I am a high school student and I have several questions how can I reach you out that′s my telegram : yorkgubler

$${sir}\:{I}\:{am}\:{a}\:{high}\:{school}\:{student}\:{and}\:{I}\:{have}\:{several}\:{questions}\:{how}\:{can}\:{I}\:{reach}\:{you}\:{out} \\ $$$${that}'{s}\:{my}\:{telegram}\::\:{yorkgubler} \\ $$

Commented by Mastermind last updated on 14/May/23

Thank you so much

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

Answered by Rajpurohith last updated on 27/May/23

Let R⊂S⊂R, ∀x∈S ,∃r∈R such that x≤r. ⇒sup(R)≤sup(S),this is a standard result. now suppose sup(R)<sup(S) ⇒∃ t∈R such that sup(R)<t<sup(S) clearly t∈S. By hypothesis ∃ r∈R such that sup(R)<t≤r<sup(S) ⇒sup(R)<r for some r ∈R a contradiction. hence our supposition is wrong. ⇒sup(S)=sup(R)

$${Let}\:{R}\subset{S}\subset\mathbb{R},\:\forall{x}\in{S}\:,\exists{r}\in{R}\:{such}\:{that}\:{x}\leqslant{r}. \\ $$$$\Rightarrow{sup}\left({R}\right)\leqslant{sup}\left({S}\right),{this}\:{is}\:{a}\:{standard}\:{result}. \\ $$$${now}\:{suppose}\:{sup}\left({R}\right)<{sup}\left({S}\right) \\ $$$$\Rightarrow\exists\:{t}\in\mathbb{R}\:{such}\:{that}\:{sup}\left({R}\right)<{t}<{sup}\left({S}\right) \\ $$$${clearly}\:{t}\in{S}. \\ $$$${By}\:{hypothesis}\:\exists\:{r}\in{R}\:{such}\:{that}\: \\ $$$${sup}\left({R}\right)<{t}\leqslant{r}<{sup}\left({S}\right) \\ $$$$\Rightarrow{sup}\left({R}\right)<{r}\:{for}\:{some}\:{r}\:\in{R} \\ $$$${a}\:{contradiction}. \\ $$$${hence}\:{our}\:{supposition}\:{is}\:{wrong}. \\ $$$$\Rightarrow{sup}\left({S}\right)={sup}\left({R}\right) \\ $$

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