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Least upper bound - greatest lower bound duality

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone!

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?! (Crying)(Crying)

Thanks!
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello everyone!

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?! (Crying)(Crying)

Thanks!
When we say that $B$ is bounded below, I think its implicit that we mean "B is bounded below in S", that is, in other words, "there is an element in S which is a lower bound to B". For otherwise it would make no sense. We have no idea what is outside S. No order relation is defined which compares elements outside S with elements in S. This is a completely abstract setting. Similarly when we say "Let L be the set of all lower bounds of B", (I think) we mean "Let L be the set of all the lower bounds of B which are in S".
It automatically follows that $L\subseteq S$.