I hope this question is not too elementary for math overflow… My knowledge on set theory is very sketchy so it will seem very simple to a working mathematician

It’s a subtle thing I only realized today: construction of free groups. The standard construction gets equivalence class of all words over a set X. Let X={a,b,c,(a,b)}. Now the words of X is the union of all the X^n for all natural numbers n. Consider the word ((a,b),c). Is it an ordered pair in X^2 or ordered triple in X^3? Of course it’s both, and that’s where the problem is. In constructing the map that shows the free group has the universal property, we pick elements out of each equivalence class (which are words), and then multiply the image of the letters in the words. But where exactly does ((a,b),c) get mapped to, if we want to factorize the function X -> integers defined by a,b,c->1 and (a,b)->10?

If ((a,b),c) is considered as a pair, it goes to 10. If considered as a triple, it goes to 1. So the mapping is ambiguous. This gets worse of course, since an 100-tuple is a 99-tuple, 98-tuple… and so has 100 different values it might get mapped to.

nonemptyordinal. $\endgroup$ – Harry Gindi Jul 16 '10 at 16:35