P is a partial order with $\leq$, and L and U are the sets of all lower and upper bounds of A
I already know that if every nonempty subset with a lower bound has an infimum, then every nonempty subset with a greater bound has a supremum:
$\forall A\subseteq S$ [A $\neq \varnothing$ L $\neq \varnothing \Rightarrow \exists$inf(A)] $\Rightarrow$ $\forall A\subseteq S$ [A $\neq \varnothing$ U $\neq \varnothing \Rightarrow \exists$sup(A)]
From there I need to show that if every subset has an infimum then every subset has a supremum:
Assuming $\forall A\subseteq S$ $\exists$inf(A)
If A = $\varnothing$ then U = S, inf(S) exists, and is sup(A)
So as long as S isn't empty:
$\forall A\subseteq S$ [$\exists$inf(A)] $\Rightarrow$ $\forall A\subseteq S$ [U $\neq \varnothing \Rightarrow \exists$sup(A)]
If S is empty, every subset of it has no inf and no sup, so all good there.
This looks like as close as I can get, obviously if U is empty there is no sup. If U is empty is there a way to show that not every subset of S has an infimum? This is where I'm stuck, and not able to quite get to $\forall A \subseteq S$ [$\exists$inf(A) $\iff \exists$sup(A)] ;_;
Edit (got it!):
If U is empty, then inf(U) = greatest(S), then every element of A is less than inf(U), so inf(U) is an upper bound of A and U $\neq \varnothing$. Assuming there are no upper bounds of A is in contradiction with the assumption that A has an infimum, so U must not be empty, and therefore A has a supremum :>