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 :>

  • 2
    $\begingroup$ just notice that the supremum of the empty set is the least element in the poset and that the infimum of the empty set is the greatest element in the poset. $\endgroup$ – Ittay Weiss Jul 31 '13 at 9:30

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