I'm not quite understanding the hypothesis behind Rudin's Theorem 1.11 (Other previous threads on 1.11 didn't have the clarification I was looking for).

Theorem: "Suppose S is an ordered set with the least-upper-bound property, B is a subset of S, B is not empty, and B is bounded below. Let L be the set of all lower bounds of B. Then A = sup(L) exists in S... (and some other conclusions)."

My question: Take B = the set of all elements in the sequence {1/(2^n)}, and let S be all positive reals. A lower bound of B is 0. Then 0 = sup(L) = inf(B), doesn't exist in B or S. I understand that B does not contain a lower bound in S in this case, but Rudin only says "bounded below", which it is in X = R, for example. And I believe my S is both ordered and has the LUB property. What's the refutation here?

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    $\begingroup$ In your example, $B$ is not bounded below in $S$. $\endgroup$ – Michael Biro Mar 20 '16 at 1:49
  • $\begingroup$ ^ I mentioned that I knew that in the last P. My question is: why not let X=R, so that B bounded below in X. How do we know Rudin intended for B bounded below in S, as he only said bounded below". $\endgroup$ – Harambe17 Mar 20 '16 at 1:53
  • $\begingroup$ Since the ordered set specified is $S$, the intention was "bounded below in $S$". $\endgroup$ – Michael Biro Mar 20 '16 at 1:58
  • $\begingroup$ Ok thank you. It seems like that has to be it, but I was thrown off bc he never stated it explicitly. $\endgroup$ – Harambe17 Mar 20 '16 at 2:01

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