Qhov siab tshaj ntawm ib txheej yog nws qhov tsawg kawg nkaus sab sauv thiab qhov infimum yog nws loj tshaj sab sauv. Lus Txhais 2.2. Piv txwv tias A ⊂ R yog cov lej tiag. Yog M ∈ R yog ib sab qaum ntawm A xws li M ≤ M′ rau txhua lub sab sauv M′ ntawm A, ces M yog hu ua lub siab tshaj ntawm A, txhais tau tias M=sup A.
Koj pom qhov zoo tshaj plaws ntawm kev ua haujlwm li cas?
Txhawm rau nrhiav qhov zoo tshaj plaws ntawm ib qho kev ua haujlwm sib txawv yog qhov teeb meem yooj yim. Piv txwv tias koj muaj y=f(x): (a, b) rau R, ces suav cov derivative dy/dx. Yog tias dy/dx>0 rau tag nrho x, ces y=f(x) nce thiab sup ntawm b thiab inf ntawm a. Yog tias dy/dx<0 rau tag nrho x, ces y=f(x) poob qis thiab qhov sup ntawm a thiab inf ntawm b.
Dab tsi yog qhov tseem ceeb ntawm txoj haujlwm?
Lub supremum (abbreviated sup; plural suprema) ntawm ib feem ntawm ib feem ntawm kev txiav txim yog qhov tsawg tshaj plaws nyob rau hauv uas yog ntau dua los yog sib npaug rau tag nrho cov ntsiab ntawm yog hais tias xws li ib tug muaj nyob. Yog li ntawd, lub supremum kuj raug hu ua qhov tsawg kawg nkaus sab sauv (los yog LUB).
Tsuas yog 1 N?
Yog koj pib ntawm n=1, koj tau 1 + 1/1 + 1/1=3, thiab qhov no yog qhov siab tshaj plaws koj yuav tau, vim txhua tus n > 1 muab rau peb tsawg dua 3. Vim tias koj tsis tuaj yeem tau ntau tshaj 3, tab sis koj tuaj yeem tau 3, nws yog ob qho tib si qhov siab tshaj thiab siab tshaj. For infimum, zaj dab neeg txawv.
Koj ua pov thawj Supremum thiab Infimum ntawm ib txheej li cas?
Ib yam li ntawd, muab ib lub bounded teeb S ⊂ R, tus lej b yog hu ua tusinfimum los yog loj tshaj qis khi rau S yog cov nram qab no tuav: (i) b yog ib tug qis khi rau S, thiab (ii) yog c yog ib tug qis khi rau S, ces c ≤ b. Yog tias b yog qhov zoo tshaj plaws rau S, peb sau tias b=sup S. Yog hais tias nws yog infimum, peb sau tias b=inf S.
