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I am reading Baby Rudin on my own to get prepared for the analysis class that I will be taking in the fall. I got up to this theorem, and I was wondering if someone can clarify me the proof of this (i.e. Thm 1.11, thefirsttheorem).

1.11 Theorem:Suppose S is an ordered set with the lub property, B [tex]\subset[/tex] S, B is not empty, and B is bounded below. Let L be the set of all lower bounds of B. Then, [tex]\alpha[/tex] = sup L exists, and [tex]\alpha[/tex] = inf B. In particular, inf B exists in S.

(The Part of) Proof:Since B is bounded below, L is not empty. Since L consists of exactly those y [tex]\in[/tex] S, which satisfy the inequality y [tex]\leq[/tex] x for every x [tex]\in[/tex] B, we see thatevery x [tex]\in[/tex] B is an upper bound of L.Thus L is bounded above. Our hypothesis is abound S implies therefore that L has a supremum in S; call it [tex]\alpha.[/tex]

The part I underlined is where I want to be clarified. I understand that S has an lub property, and that implies that if E [tex]\subset[/tex] S, E is not empty, and E is bounded above, then sup E exists in S (EDIT: Originally it said "... then sup E exists in E, until rasmhop pointed out the misprint.). While I understand why L is bounded above, I am not clear on how this proves the existence of sup L. My guess is that we're concerned about the set K [tex]\subset[/tex] S [tex]\cap[/tex] L, and by definition sup K exists, and thus sup L exists, but my guess could be wrong.

Please let me know if you can help me out. Notice that I only put the part of the proof here since I believe the rest is irrelevant, but please let me know if you want to read the whole proof--I can always edit it later.

Thanks.

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# Question from a theorem in Baby Rudin (Re: Least-Upper-Bound Property)

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