
theorem Th9:  :: variant of SUBSET_1:46 or CARD_1:65
for X being set, s being Subset of X
 st s is 1-element
  ex x being set st x in X & s = {x}
proof
 let X be set, s be Subset of X;
 assume s is 1-element;
 then s is trivial non empty;
 then consider x being Element of s such that
A1: s = {x} by SUBSET_1:46;
 take x;
   x in s by A1;
 hence x in X;
 thus s = {x} by A1;
end;
