Partial Orders: "Minimum property" iff "Maximum property"? 
Given a partially ordered set $(X,\leq)$, is it true that
  $$
\forall P\subseteq X:P\neq\emptyset\text{ and }P\text{ is bounded below }\to P\text{ has a minimum}\\
\updownarrow\\
\forall P\subseteq X:P\neq\emptyset\text{ and }P\text{ is bounded above }\to P\text{ has a maximum}
$$

If we replace minimum and maximum by infimum and supremum respectively, then a proof has been given here. However, in that proof, I don't see how one would infer $g\in P$ from $g\in B$.

In the linearly ordered set $X = \{-\frac{1}{n}: n=1,2,\ldots\} \cup \{0\}$ (in the order inherited from the reals) we have that $X$ satisfies the upper condition (check this), but not the lower, because the left part has an upper bound $0$, but no maximum.
So the properties are not equivalent. Use $X = \{0\} \cup \{\frac{1}{n}: n=1,2,\ldots\}$ for the other implication, of course.
