Yes, that was exactly my line of thinking. One minor nitpick: f(x)=1/x maps [-1,1] to [-inf,-1]u[+1,+inf] ... that is, 1 and -1 are in both sets and mapped ont

Yes, that was exactly my line of thinking. One minor nitpick: f(x)=1/x maps [-1,1] to [-inf,-1]u[+1,+inf] ... that is, 1 and -1 are in both sets and mapped onto themselves. However, this detail doesn't affect the argument much.