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#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

There's a point I didn't get in Rudin's theorem 1.11 that says:

Suppose S is an ordered set with the LUB property, and B $\subset$ S, B is not empty and B is bounded below. Let L be the set of lower bounds of B. Then a = sup L exists in S, and a - inf B. In particular inf B exists in S.

Now the proof states that L is not empty (which is fine by me) and bounded above (fine by me too), hence "our hypothesis about S implies therefore that L has a supremum in S;"

Who said that L $\subset$ S so we can make this strong conclusion?!

Thanks!