# 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.

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!!

• What if $X=\{\,1,1/2,1/3,1/4,\dots\,\}$? – Gerry Myerson Mar 23 at 3:00
• @GerryMyerson It has both: every subset bounded below is finite. – Gae. S. Mar 23 at 3:05

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.

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.