Question

how to give explicit bijection between poweret(Z) and poweset(N)?

Answer 1

it suffices to show there is a bijection between Z and N. Explicitly, this can be given by N(z)=2*z+1, if z is positive, and -2*z if z is negative.

Answer 2

this is because powerset is a dependent upon the # of unique elements and not the value of each unique element.

Answer 3

*the number of elements in a powerset

Answer 4

i suppose you can say an element of powerset(z) containing {a,b,c,d,...}=an element of powerset(n) containing {n(a),n(b),n(c),n(d),...}, where n(z)=2*z+1 if z>0 and -2*z is z<0.

Answer 5

why do you use n(z)=2*z+1

Answer 6

my notation is probably wrong; the general idea is that given some integer z, the corresponding natural number n will be equal to 2*|z|+sgn(z), where sgn(z) is 1 if z is positive and 0 otherwise.

for example, {-3,-2,-1,0,1,2,3} corresponds to {6,4,2,1,3,5,7}.