Is every one-one function from N to itself onto? In other words, does the pigeon-hole principle hold for N itself?
Thanks for the help
No. Consider the function f:N->N defined by f(n) = 2n, or the function g:N->N defined by g(n) = 2n + 1. Both these functions are one to one, but not unto. f only maps to the even integers in N, while g only maps to the odd integers in N