Theorem: Suppose S is an ordered set with the least-upper-bound property, B⊂S, B is not empty, and B is bounded below. Let L be the set of all lower bounds of B. Then α=supL exists in S, and α=infB.(adsbygoogle = window.adsbygoogle || []).push({});

Rudin proves that α=supL, α is an element of L and that α=infB.

For α to be sup i.e. lub it has to be in set of upper bounds. Here i quote rudin: "Since B is bounded below, L is not empty. Since L consists of exactly those y that are elemet of S which satisfy the inequality y≤x for every x from B, we see that evry x that is in B is an upper bound of L. Thus L is bounded above. Our hypothesis about S implies therefore that L has a supremum in S; call it α. If γ<α then γ is not an upper bound of L, hence γ is not an element of B. It follows that α≤x for every x in B. Thus α is an element of L. If α<β then β is not an element of of L, since α is an upper bound of L. We have shoh that α is an element of L but β is not element of L if α<β. In other words , α is lower bound of B, but β is not is not if α<β. This means that α=infB."

So since b consists of all of upper bounds of L and α is least upper bound, can i say that α is also an elemet of B?

If that is the case can i then say that maxL=supL=infB=minB?

Because α is an element of L and at the same time α=supL then maxL=α. If it were not the cas that maxL=α then there would be some other element β in L, that is greater then α. But this would contadict the fact that supL=α. Same goes for minB=α just reveresed.

Are these conclusions correct?

P.S. Why i don't have LATEX reference in editor?

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# Baby Rudin Theorem 1.11

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