Let $(P, \leq)$ be a partially ordered set, and let $A$ be a subset of $P$. The set $UB(A)$ is the set of upper bounds of $A$, where upper bound means greater than or equal to every element of $A$. Now, what happens if we form $UB(UB(A))$? I conjecture that it is either empty (if $P$ has no maximum), or it is a singleton (if $P$ does have a maximum). Is this correct?
2026-04-19 12:45:29.1776602729
Is the set of upper bounds squared either empty or a singleton?
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Your conjecture is not quite right, for two reasons.
First, you've forgotten the case when $UB(A) = \varnothing$. Then $UB(UB(A)) = P$, since every element of $P$ is an upper bound for the empty set.
Second, it's possible to get $UB(UB(A))$ to be a singleton even when $P$ has no greatest element - all you need is that $UB(A)$ has a greatest element, not all of $P$.
For an example demonstrating both behaviors, let $P = \{a,b\}$ with $a$ and $b$ incomparable. Then $UB(UB(\{a,b\})) = UB(\varnothing) = \{a,b\}$. And on the other hand, $UB(UB(\{a\}) = UB(\{a\}) = \{a\}$, even though $P$ has no greatest element.
To answer the question correctly, let's consider $UB(X)$ where $X\subseteq P$ is any upwards-closed set.
Now for an arbitrary set $A\subseteq P$, just note that $UB(A)$ is upwards-closed and apply the above to get: $$UB(UB(A)) = \begin{cases} P & \text{ if $A$ has no upper bound}\\ \varnothing & \text{ if $A$ has an upper bound but no greatest upper bound}\\ \{g\} & \text{ if $A$ has a greatest upper bound $g$}.\end{cases}$$