[2.4] Exercise 8
For a given set \(N\), there can be at most one set \(M\) such that \(N,M\) is an Euler pair. Why? Think backward: "If there were two different such sets, \(M\) and \(M'\), then there would have to be some smallest integer \(n\) that lies in one set but not in the other" What does this mean for \(p(n \mid \text{distinct parts in} M)\) compared to \(p(n \mid \text{distinct parts in }M')?\)

Proof

By the hint, suppose there were two different sets \(M\) and \(M'\). Then without loss of generality, there must be a smallest integer \(n\) that exists in \(M\) and not \(M'\). Now consider partitioning \(n\) into distinct parts in \(M\). Since \(n \in M\), then \(n\) is a valid partition of \(n\) into distinct parts of \(M\). But since \(n \notin M'\), then \(n\) is not a partition of \(n\) into distinct parts of \(M'\). Moreover, since \(n\) is the smallest integer where it exists in \(M\) but not \(M'\), then the set of all elements less than \(n\) in \(M\) is equal to the set of all elements less than \(n\) in \(M'\). Hence, the number of partitions of \(n\) using smaller elements than \(n\) in \(M\) is the same as the number of partitions of \(n\) in \(M'\) using smaller elements than \(n\) . But this means that both sets differ in exactly one partition. This means that

$$ \begin{align} p(n \mid \text{distinct parts in } M) \neq p(n \mid \text{distinct parts in } M') \end{align} $$

But we know \((N,M)\) is an Euler pair so

$$ \begin{align} p(n \mid \text{parts in } N) = p(n \mid \text{distinct parts in } M) \end{align} $$

and Similarly \((N,M')\) is an Euler pair so

$$ \begin{align} p(n \mid \text{parts in } N) = p(n \mid \text{distinct parts in } M') \end{align} $$

This is a contradiction. Therefore, we must have that \(M' = M. \blacksquare\)

References