Equivalence of two statements on an arbitrary partially ordered set $(A, <)$

Question

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 $C\subset A$ 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 $S\in D$. 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!

Answer 1

Suppose $S\not\in D$. Then $S$ is not a lower bound for $C$ and so there exists $c\in C$ such that $c\lt S$. 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 $c\in C$ and any $d\in D$, $d\le c$ so $c$ is an upper bound for $D$. Since $S$ is the least upper bound, $S\le c$. Since this is true for every $c\in C$, $S$ is a lower bound for $C$ and hence, is in $D$.

Answer 2

If $\varnothing\neq C\subseteq A$ and $C$ is bounded from below, and $D:=\{d\in A\mid d\text{ is a lower bound of }C\}$ then $D$ is not empty and: $$\forall c\in C\forall d\in D\;d\leq c\tag1$$ where $d\leq c$ abbreviates $d=c\vee d<c$.

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 $c\in C$ is an upper bound of $D$ so that $s\leq c$ for every $c\in C$ (because $s$ is the least upper bound of $D$).

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

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