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In Topology by Munkres, an alternative approach to proving that "an ordered set $A$ with the greatest lower bound property satisfies the least upper bound property" (this is 14c) is suggested below:

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I know how to show 14a and 14b (Chapter 3). How can one use these parts to prove 14c?

For clarity of definitions:

The greatest lower bound property is defined as the property that for a set $S \subseteq A$, if $S$ has a lower bound in $A$, then its $\inf$ is also in $A$. The least upper bound property is defined similarly.

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Let $S\ne\emptyset$ be bounded above. Since by definition $\sup S=\min\{ x\in A\,:\, \forall s\in S,\ x\ge s\}$, you have no choice but proving that $\inf\{x\in A\,:\, \forall s\in S,\ x\ge s\}\in\{x\in A\,:\,\forall s\in S,\ x\ge s\}$ (why is this set bounded below, by the way?). Or, equivalently, that $$\forall u\in S,\ \inf\{x\in A\,:\, \forall s\in S,\ x\ge s\}\ge u$$

Assume as a contradiction that there is $u\in S$ such that $\inf\{x\in A\,:\, \forall s\in S,\ x\ge s\}<u$. But $u\in S$ is a lower bound of $\{x\in A\,:\,\forall s\in S,\ x\ge s\}$. Absurd.

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  • $\begingroup$ This works, but does it use 14a and 14b? $\endgroup$ – Muno May 16 '17 at 19:25
  • $\begingroup$ No, it does not. Perhaps the author wants you to prove that if a total order satisfies the greatest lower bound property, then it reverse does too. I don't see a special argument for it, though. $\endgroup$ – user228113 May 16 '17 at 19:31

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