Suprema and infima on a partially ordered set This is Exercise 6 from page 28 of Analysis I by Amann and Escher. I have searched for "supremum union" on this site and there are a number of similar questions but they all seem to assume some extra structure that I can't justify.
Exercise:



Comments:
I have made almost no progress. Some of the similar questions on StackExchange involve the real numbers, or proofs that for example $\sup(A \cup B) = \max \{ \sup(A), \sup(B) \}$, neither of which are applicable here. Some other questions involve proving that certain sets are bounded, which doesn't seem relevant when all the suprema and infima are assumed to exist.
One thing that I feel is paralyzing me is that because this is a partially ordered set, I can't necessarily compare any two elements. Since $A$ is bounded above I can presumably use the relation $\leq$ on an upper bound $s$ of $A$ and every $a \in A$. But I can't, for example, necessarily compare $s$ with an upper bound $t$ of $B$.
I'm starting to think that, for part (a) at least, I should pursue a strategy along the lines of showing that $\sup(A \cup B) \leq \sup \{ \sup(A), \sup(B) \}$ and then $\sup(A \cup B) \geq \sup \{ \sup(A), \sup(B) \}$. Is that reasonable?
I appreciate any help.
 A: Your final paragraph outlines a perfectly reasonable strategy for (a). So let's do that. And remember the defining property of $\sup$ (and analogously for $\inf$): It is the least upper bound, meaning all other upper bounds are larger. So any inequality of the form $\sup(X) \leq t$ is best proven by showing that $t$ is an upper bound for $X$.

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*$\sup(A\cup B)\leq \sup\{\sup(A), \sup(B)\}$: Let $x = \sup\{\sup(A), \sup(B)\}$, and take an element $a\in A\cup B$. Either $a\in A$, which means $a\leq \sup(A)\leq x$, or $a\in B$, which means $a\leq \sup(B)\leq x$. Either way we get $a\leq x$. This means $x$ is an upper bound for $A\cup B$.

*$\sup(A\cup B)\geq \sup\{\sup(A), \sup(B)\}$: Clearly, by using (b) ("larger set means larger $\sup$"), we have $\sup(A\cup B)\geq \sup(A)$, and just as clearly $\sup(A\cup B) \geq \sup(B)$. Thus $\sup(A\cup B)$ is an upper bound for $\{\sup(A), \sup(B)\}$.

The $\inf$ proof is entirely analogous, except we flip all the inequality signs, and change "upper" into "lower" (same goes for (b) and (c) as well).
Now for (b). This time there is only one inequality. It's rather simple to prove, though, using the defining property: We have that $\sup(B)$ is an upper bound for $B$, so it must be an upper bound for $A$.
The proof for (c) is very similar to half the proof for (a): again by using (b), clearly $\sup(A\cap B)\leq \sup(A)$, and just as clearly, $\sup(A\cap B)\leq \sup(B)$. Thus $\sup(A\cap B)$ is a lower bound for $\{\sup(A), \sup(B)\}$. (One could go the other way too, showing that $\inf\{\sup(A), \sup(B)\}$ is an upper bound for $A\cap B$. That would be more akin to the other half of the proof of (a). This time there isn't equality, because the two halves show the same inequality, rather than opposite inequalities the way they did in (a).)
Finally, we have (d). They give one hint as to what you could look at, but I like to do it simpler. Let us say our partially ordered set has three elements, two are incomparable, and the third is larger than both (the power set on a set of two elements yields a similar example). Then let $A$ and $B$ each consist of one of the two incomparable elements. Then the set $\{\sup(A), \sup(B)\}$ has no maximum as its elements are incomparable, by design.
