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In the baby Rudin there's a theorem that every ordered set with the least-upper bound property also has the greatest-lower-bound property. I wonder if the converse is true?

I tried to prove it and it looks exactly the same as the least upper bound $\rightarrow$ greast lower bound theorem.

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  • $\begingroup$ it should look the other way round, but yes, the idea is the same. $\endgroup$ – Thomas Dec 22 '14 at 19:19
  • $\begingroup$ Yes, they are equivalent. The argument (with appropriate changes of inequality symbols and sup and inf switched) is basically the same. $\endgroup$ – Cameron Williams Dec 22 '14 at 19:20
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Let it be that order $(X,<)$ has the greatest upper bound property.

Then order $(X,>)$ has the least upper bound property and (as you said) consequently $(X,>)$ has the greatest upper bound property.

The last statement means exactly that order $(X,<)$ has the least upper bound property.

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