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I think is a very dumb question, but i failed to find in the books if this statement is true. One can easily proove that if a ordered set S has the Least Upper Bound Property, then every non empty subset of S, that is bounded from below, has a Greatest Lower Bound in S.

But is equally true, that, if a ordered set M has the Greatest Lower Bound Property, then every non empty subset of M, that is bounded from above, has a Least Upper Bound in M?

Thank's a lot!

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    $\begingroup$ Yes, they imply each other. To prove it, simply use $$\sup A= \inf \{ \mbox{upper bounds of } A \}$$ and viceversa $$\inf A= \sup \{ \mbox{lower bounds of } A \}$$ $\endgroup$ – Crostul Mar 16 '17 at 18:47
  • $\begingroup$ Awesome! Thanks! $\endgroup$ – S. Cow Mar 16 '17 at 19:00

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