Let us show that "W=\\{A\\subset S\\ |\\ A\\text{ is finite }\\}".
Let us prove using mathematical induction on the order of the elements of "W".
Base case:
n=0: "\\emptyset" is a unique subset of "S" of order 0, and by defenition of "W", "\\emptyset\\in W".
n =1: for each "x\\in S" the singleton "\\{x\\}=\\{x\\}\\cup\\emptyset\\in W", and thus "W" contains all singletons.
Inductive step:
Assume that "W" contains all subsets "A\\subset S" of cardinality "|A|=k", and prove that it is also contains all subsets od cardinality "k+1". Indeed, let "B=\\{x_1,...,x_k,x_{k+1}\\}" arbitrary subsets of cardinality "k+1". Then by assumption, "A=\\{x_2,....x_{k+1}\\}\\in W," and consequently, "B= \\{x_{1}\\}\\cup A\\in W".
Conclusion:
Therefore, by Mathematical Induction "W=\\{A\\subset S\\ |\\ A\\text{ is finite }\\}".
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