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I have come across this problem in my discrete mathematics class and I have no clue how to go about it since I haven't dealt with upper bounds before in sets. If anyone could help me out, I'd greatly appreciate it.

In the special case S = {1, 2, 3, 4}, there exists two sets A,B that are elements of P(S) such that {A,B} has no upper bounds.

2 Answers 2

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You need to specify your ordering. It seems safe to assume that you mean membership (i.e. $A \leq B \Leftrightarrow A \subseteq B$). In that case it seems like there are many sets $A,B$ that satisfy the condition that $A,B \in P(S)$ and $\{A,B\}$ has no upper bounds. For instance $A = \{1\}$ and $B = \{2\}$.

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What order are you considering? The natural one would be to order the subsets by inclusion. In this case, $S$ is an upper bound for $P(S)$. But if $A$ and $B$ are elements of $P(S)$, then $\{A,B\}$ looks something like $\{\{2\},\{3\}\}$ and is not in $P(S)$