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Question-192297




Question Number 192297 by Mastermind last updated on 14/May/23
Answered by witcher3 last updated on 14/May/23
let a=supA,b=SupB⇒∀(x∈Aety∈B)  x+y≤a+b⇒sup(A+B)≤a+b  let M=sup(A+B)  ∀ε>0 ∃x∈A,y∈B such a−ε<x≤a  b−ε<y≤b  ⇒a+b−2ε<x+y∈A+B≤a+b  ⇒∀ε>0 ∃S=x+y∈A+B  such a+b−ε<x+y  ⇒sup(A+B)≥a+b=sup(A)+sup(B)  ⇒sup(A+B)=sup(A)+sup(B)
$$\mathrm{let}\:\mathrm{a}=\mathrm{supA},\mathrm{b}=\mathrm{SupB}\Rightarrow\forall\left(\mathrm{x}\in\mathrm{Aety}\in\mathrm{B}\right) \\ $$$$\mathrm{x}+\mathrm{y}\leqslant\mathrm{a}+\mathrm{b}\Rightarrow\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right)\leqslant\mathrm{a}+\mathrm{b} \\ $$$$\mathrm{let}\:\mathrm{M}=\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\forall\epsilon>\mathrm{0}\:\exists\mathrm{x}\in\mathrm{A},\mathrm{y}\in\mathrm{B}\:\mathrm{such}\:\mathrm{a}−\epsilon<\mathrm{x}\leqslant\mathrm{a} \\ $$$$\mathrm{b}−\epsilon<\mathrm{y}\leqslant\mathrm{b} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}−\mathrm{2}\epsilon<\mathrm{x}+\mathrm{y}\in\mathrm{A}+\mathrm{B}\leqslant\mathrm{a}+\mathrm{b} \\ $$$$\Rightarrow\forall\epsilon>\mathrm{0}\:\exists\mathrm{S}=\mathrm{x}+\mathrm{y}\in\mathrm{A}+\mathrm{B} \\ $$$$\mathrm{such}\:\mathrm{a}+\mathrm{b}−\epsilon<\mathrm{x}+\mathrm{y} \\ $$$$\Rightarrow\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right)\geqslant\mathrm{a}+\mathrm{b}=\mathrm{sup}\left(\mathrm{A}\right)+\mathrm{sup}\left(\mathrm{B}\right) \\ $$$$\Rightarrow\mathrm{sup}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{sup}\left(\mathrm{A}\right)+\mathrm{sup}\left(\mathrm{B}\right) \\ $$$$ \\ $$
Commented by Mastermind last updated on 14/May/23
I do really appreciate    Thank you man
$$\mathrm{I}\:\mathrm{do}\:\mathrm{really}\:\mathrm{appreciate} \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{man} \\ $$
Answered by witcher3 last updated on 14/May/23
(2),a_n =arctan(n),b_n =arctan(−n)  a_n +b_n =0  supa_n +supb_n =π
$$\left(\mathrm{2}\right),\mathrm{a}_{\mathrm{n}} =\mathrm{arctan}\left(\mathrm{n}\right),\mathrm{b}_{\mathrm{n}} =\mathrm{arctan}\left(−\mathrm{n}\right) \\ $$$$\mathrm{a}_{\mathrm{n}} +\mathrm{b}_{\mathrm{n}} =\mathrm{0} \\ $$$$\mathrm{supa}_{\mathrm{n}} +\mathrm{supb}_{\mathrm{n}} =\pi \\ $$$$ \\ $$
Commented by Mastermind last updated on 14/May/23
Thank you my BOSS  is that all?
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{BOSS} \\ $$$$\mathrm{is}\:\mathrm{that}\:\mathrm{all}? \\ $$

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