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In Rudin's Principles of Mathematical Analysis, we have the following theorem early on:

Theorem 1.11. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $$ \alpha = \sup L $$ exists in $S$, and $\alpha = \inf B$.

In particular, $\inf B$ exists in $S$.

Proof. Since $B$ is bounded below, $L$ is not empty. Since $L$ consists of exactly those $y \in S$ which satisfy the inequality $y \leq x$ for every $x \in B$, we see that every $x \in B$ 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 $\alpha$.

If $\gamma < \alpha$ then (see Definition 1.8) $\gamma$ is not an upper bound of $L$, hence $\gamma \not\in B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus, $\alpha \in L$.

If $\alpha < \beta$ then $\beta \not\in L$, since $\alpha$ is an upper bound of $L$.

We have shown that $\alpha \in L$ but $\beta \not\in L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta$ is not if $\beta > \alpha$. This means that $\alpha = \inf B$.

Why it important that $S$ has the least upper bound property? Going through the proof, I don't see what would change if $S$ didn't have the LUBP.

Additionally, Rudin says beforehand that LUBP implies GLBP for an arbitrary set, but all I see is that if some set $A$ in a set $S$ has LUBP, there exists another set $B$ in $S$ that has GLBP. Is that the result we wanted?

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    $\begingroup$ "Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$" it's used here $\endgroup$ – Alessandro Codenotti Sep 8 '18 at 23:28
  • $\begingroup$ The set $A$ in $S$ (mathematically $A\subseteq S$) does not have LUBP, rather the correct statement is "$A$ has LUB in $S$ and $S$ has LUBP". The theorem being proved is that if some ordered set $S$ has LUBP then it also has GLBP. $\endgroup$ – Paramanand Singh Sep 8 '18 at 23:32
  • $\begingroup$ @ParamanandSingh But I thought we only proved that B has GLBP. How does that extend to S? $\endgroup$ – Tiwa Aina Sep 8 '18 at 23:44
  • $\begingroup$ No the thing being proved is that "if $B\neq \emptyset, B\subseteq S$ and $B$ is bounded below then $B$ has a GLB (not GLBP) in $S$". This is what mean by the statement "$S$ has GLBP". $\endgroup$ – Paramanand Singh Sep 8 '18 at 23:47
  • $\begingroup$ I think you are trying to mix the term LUB with LUBP and similarly GLB with GLBP. LUBP / GLBP is a property / feature which some set $S$ may have, but LUB, GLB are members of $S$. $\endgroup$ – Paramanand Singh Sep 8 '18 at 23:49
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Note that if $S$ doesn't have the least-upper-bound property, then there is no reason why we can assert that $\sup L$ exists. That is the reason why Rudin wrote that “Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$”.

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  • $\begingroup$ Would you be able to provide an example of an $S$ without LUBP and a $B$ such that $L$ has no supremum in $S$? $\endgroup$ – Tiwa Aina Sep 8 '18 at 23:29
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    $\begingroup$ @TiwaAina Sure. $S=\mathbb{Q}$, with its usual order, and $B=\{q\in\mathbb{Q}\,|\,q^2<2\}$. Then $L=\left(-\infty,-\sqrt2\right)\cap\mathbb Q$, which has no supremum in $\mathbb Q$. $\endgroup$ – José Carlos Santos Sep 8 '18 at 23:33
  • $\begingroup$ @Tiwa $S=\Bbb R\setminus\{0\}$ and $L=(-\infty,0)$ $\endgroup$ – Alessandro Codenotti Sep 8 '18 at 23:33
  • $\begingroup$ @JoséCarlosSantos Oh, I see! One last question (this is also mentioned in my initial post): Rudin says beforehand that LUBP implies GLBP for an arbitrary set, but all I see is that if some set $A$ in a set $S$ has LUBP, there exists another set $B$ in $S$ that has GLBP. Is that the result we wanted? $\endgroup$ – Tiwa Aina Sep 8 '18 at 23:39
  • $\begingroup$ @TiwaAina Rudin proves that if every non-empty set in $S$ has a supremum, then every non-empty set in $S$ has an infimum. $\endgroup$ – José Carlos Santos Sep 8 '18 at 23:45

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