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I have a question in my homework for my math proof's class. Let S be a collection of n≥2 numerically equivalent sets. Prove that these sets can be shown to be numerically equivalent (same cardinality) by means of n-1 bijective functions between pairs of sets in S.

My problem here is the "n-1 bijective funtions" part. Does this mean that if S has two numerically equivalent sets, then there is at least one function between them?

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    $\begingroup$ What does "numerically equivalent" mean? $\endgroup$ – Trevor Wilson Nov 19 '13 at 4:10
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It means that if $S$ has two numerically equivalent sets, then it takes only one bijection to prove it.

More generally, if $S$ is a set of $n\ge 2$ numerically equivalent sets, it is saying that with $n-1$ bijections, we can show that all of the members of $S$ are numerically equivalent. We can show it by exhibiting a bijection between every pair of elements of $S$--for a total of $\frac{n(n-1)}2$ bijections--but the claim says that it is enough to have $n-1$, if we choose them carefully.

Hint: Show that a composition of finitely-many one-to-one/onto functions is again one-to-one/onto (respectively).

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