for all y>0 there exists nεN SUCH THAT n>y
SINCE A is not empty and also a subset of the natural Nos N, then according to the well ordering principal of natural Nos :
there exists a kεA AND .
But since kεA ,THEN y<k. ALSO k-1<k.
If now y<k-1 ,then (k-1)εN which implies a contradiction ,hence :
For part (b).
Assume that the set A HAS two minimums k,m both belonging to A
But since m is a minimum , ...........(1)
Also since k a minimum , .............(2)
From (1) and(2) we conclude k=m ,therefor k is unique
Do not forget that a minimum of a set is an infemum that belongs to the set.
If you do not understand part (b) i can give a more detailed proof