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Proposition: Let X be a linearly ordered set with order <. Then X has the least upper bound property iff (if and only if) X has the greatest lower bound property.

I know that if I Let X be a linearly ordered set with order <. We say that X has the least upper bound property iff every nonempty subset A $\subset$ X that is bounded above has a least upper bound. Similarly, we say that X has the greatest lower bound property iff every nonempty subset A $\subset$ X that is bounded below has a greatest lower bound. However, I dont know how to show or relate both the things.

I need to show that if X has LUB property, then X has GLB property and the other way. Can someone help me with a complete proof of this? Let me know if you have questions!!

Appreciate your patience and time.

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    $\begingroup$ What if $X=\{\,1,1/2,1/3,1/4,\dots\,\}$? $\endgroup$ – Gerry Myerson Mar 23 at 3:00
  • $\begingroup$ @GerryMyerson It has both: every subset bounded below is finite. $\endgroup$ – Gae. S. Mar 23 at 3:05
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Assume GLB.
Let A be a not empty, bounded above subset.
Let B be the set of all upper bounds of A.
Show sup A = inf B. (LUB A = GLB B)

LUB implies GLB is simply the order dual of above.

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Given and ordered set $(X,\le)$ and $A\subseteq X$, call $$L_A=\{x\in X\,:\, \forall y\in A, x\le y\}\\U_A=\{x\in X\,:\, \forall y\in A, y\le x\}$$

and notice that $A\subseteq U_{L_{A}}$.

Let $(X,\le )$ be an ordered set with the least upper bound property and let $S\subseteq X$ be non-empty and bounded below: $L_S$ is non empty by hypothesis and, by $S\subseteq U_{L_{S}}$, it is bounded above. Consider therefore $\sup L_S=\min U_{L_S}$. Since $U_{L_S}\supseteq S$, we have that $\min U_{L_S}\le y$ for all $y\in S$, and therefore $\sup L_S\in L_S$. Therefore, $\sup L_S$ is actually $\max L_S$ as desired.

On the other hand, let $(X,\le )$ be an ordered set with the gratest lower bound property. Then $(X,\ge)$ is an ordered set with the least upper bound property, and by the previous lemma $(X,\ge)$ has the greatest lower bound property. Therefore $(X,\le)$ has the least upper bound property.

Notice that the order being total is unnecessary.

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