Vrftjkry

Let $(A, <)$ a arbitrary partially ordered set. ( $<$ is irreflexive and transitive)

These two statements are equivalent:

Every nonempty subset of $A$ that is bounded from above has a supremum in $A$

Every nonempty subset of $A$ that is bounded from below has an infimum in $A$

My attempt:

Let the first statement be true and I will try to prove the second follows.(I suspect the second direction will be pretty much the same)

Let $Csubset A$ be an arbitrary subset of $A$ that is bounded from below.
Let $D$ denote the set of all lower bounds of $C$.

So we have $forall c in C, forall d in D, d < c$.

I have to prove that $D$ has the biggest element(definition of infimum).
I know that $D$ is a set that is bounded from above, then it follows that $D$ has a supremum, let's denote it by $S$.

Now, I know, $(forall d in D)( d < S lor d = S)$. But I don't know how to prove that $Sin D$. Because if $S notin D$ then $S$ doesn't have to be comparable to any $c in C$ since the set is partially ordered, then $S$ is definitely not the infimum of $C$ which I think it should be.

Thanks in advance!

Suppose $Snotin D$. Then $S$ is not a lower bound for $C$ and so there exists $cin C$ such that $clt S$. This is impossible because all elements of $c$ are upper bounds for $D$ so $S$ would not be the least upper bound for $D$. This is a contradiction.

Edit: There is a slight problem with this proof due to the fact that the order is a partial order. Thank you to drhab for pointing it out. The fix is simple enough and I can do it without deriving a contradiction.

For any $cin C$ and any $din D$, $dle c$ so $c$ is an upper bound for $D$. Since $S$ is the least upper bound, $Sle c$. Since this is true for every $cin C$, $S$ is a lower bound for $C$ and hence, is in $D$.

If $varnothingneq Csubseteq A$ and $C$ is bounded from below, and $D:={din Amid dtext{ is a lower bound of }C}$ then $D$ is not empty and:
$$forall cin Cforall din D;dleq ctag1$$
where $dleq c$ abbreviates $d=cvee d<c$.

Since $D$ is not empty it will have a least upper bound $s$ (I prefer a small $s$ here and preserve capitals for sets).

It is our aim to prove that $s$ is greatest lower bound of $C$.

$(1)$ tells us that every $cin C$ is an upper bound of $D$ so that $sleq c$ for every $cin C$ (because $s$ is the least upper bound of $D$).

This proves that $s$ is a lower bound of $C$ or equivalently that $sin D$.

Next to that we have $dleq s$ for every $din D$ (i.e. for every lower bound of $C$) so $s$ can be classified as greatest lower bound of $C$.