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 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)

R⊆S⇒x=Sup(R)⩽sup(S)=y∀r∈Rr⩽y,∀ϵ>0∃s∈Ssuchy−ϵ<s⩽ybydefinition∃r∈Rr⩾s⇒∀ϵ>0y−ϵ⩽rϵ=1n⇒∀n∈Ny−1n<r⇒s⩾y⇒sup(S)⩽Sup(R)⇔sup(R)=Sup(S)

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

sirIamahighschoolstudentandIhaveseveralquestionshowcanIreachyououtthat′smytelegram:yorkgubler

Commented by Mastermind last updated on 14/May/23

Thank you so much

Thankyousomuch

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)

LetR⊂S⊂R,∀x∈S,∃r∈Rsuchthatx⩽r.⇒sup(R)⩽sup(S),thisisastandardresult.nowsupposesup(R)<sup(S)⇒∃t∈Rsuchthatsup(R)<t<sup(S)clearlyt∈S.Byhypothesis∃r∈Rsuchthatsup(R)<t⩽r<sup(S)⇒sup(R)<rforsomer∈Racontradiction.henceoursuppositioniswrong.⇒sup(S)=sup(R)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com