S = {xIx= 1/2^n，n∈N*}

The supremum of is 1/2, and the infimum is 0. The supremum is contained in S and is not proved; Only the latter, in two steps:

1) First of all, any x∈S has x >；; 0;

2) secondly, for any ε >; 0(ε& lt； 1), all with n = [ln (1/ε)/LN2]+1,so,

x = 1/2^n = 1/2^{[ln( 1/ε)/ln2]+ 1}≥2^{ln( 1/ε)/ln2} =ε，

According to the definition of infimum, we know that infS=0.