I'm reading A Book of Set Theory by Charles C. Pinter. I don't know how to solve exercise 4.3.11.
Let $A$ be a partially ordered class. Prove the following:
a) If every subclass of $A$ has a $\sup$ and an $\inf$ in $A$, then $A$ has a least element and a greatest element. [Hint: Use 4.27.]
b) The following two statements are equivalent: Every subclass of $A$ has a $\sup$; every subclass of $A$ has an $\inf$.
Part a) seems straightforward: $A$ is a subclass of itself so $\sup A$ and $\inf A$ exist and must be its greatest and lowest elements respectively.
However I'm confused with b). Isn't it obviously false? Consider the partial order defined on the class $A=\{1,2,3\}$ represented in this diagram or equivalently given by the graph $G=\{(1,1),(1,2),(1,3),(2,2),(3,3)\}$.
Then any subclass of $A$ has an infimum but $A$ itself does not have a supremum, right?
What am I missing?
In your example not every subclass of $A$ has an infimum. The set of lower bounds of the empty set is {$1,2,3$} and this does not have a greatest element.
To show part 2), suppose every subset of $A$ has a sup. Let $B$ be a subset of $A$. We will show $B$ has an inf by showing that the sup of the set of lower bounds of $B$ (which we know to exist by our assumption), $L(B)$, is the inf of $B$.
Suppose $B \ne \phi.$ Let $x=$ Sup $L(B)$ and suppose $y\in B$. Let $z\in L(B)$. Then we know $z\le y$ as $z$ is a lower bound for $B$. Hence $y$ is an upper bound for $L(B)$. By definition of Sup, we know $x \le y$, as $x$ is the least upper bound for $L(B)$. Hence as $y$ was arbitrary, $x$ is a lower bound for $B$. Now suppose $w \in L(B)$. Then $w \le x$ as $x$ is an upper bound of $L(B)$. Hence $x$ is the greatest lower bound of $B$.
If $B = \phi$ then we must show $ \phi$ has an inf. Or the set of lower bounds of $ \phi$ has a greatest element. Or $A$ has a greatest element. But we know from our assumption that $A$ has a sup in $A$. Therefore $A$ has a greatest element.
The reverse direction follows a similar argument.