Talk:Set Equivalence Less One Element

This can't be done without LEM either, can it? This construction of $g$ requires dividing into cases based on whether an element of $S$ is or is not equal to $f^{-1}(b)$. If however you construct $g$ by taking the intersection of $f$ with the product of the set differences and adding $(f^{-1}(b),f(a))$, then you need LEM to prove that $g$ is a mapping on $S \setminus \{ a \}$. In the finite case, I imagine proof by cases will do the trick. --Dfeuer (talk) 20:22, 12 March 2013 (UTC)