Why?
Write \\(S = \lbrace x_1, \ldots, x_n \rbrace\\).
Given this ordering for the elements, each subset \\(A\\) corresponds to a vector of \\(n\\) bits, with the ith bit saying whether or not \\(x_i\\) belongs to \\(A\\).
So each subset can be identified by an n-digit sequence in base 2.
Since there are \\(2^n\\) of these sequences, that means the power set has size \\(2^n\\).
Write \\(S = \lbrace x_1, \ldots, x_n \rbrace\\).
Given this ordering for the elements, each subset \\(A\\) corresponds to a vector of \\(n\\) bits, with the ith bit saying whether or not \\(x_i\\) belongs to \\(A\\).
So each subset can be identified by an n-digit sequence in base 2.
Since there are \\(2^n\\) of these sequences, that means the power set has size \\(2^n\\).
Version 2.1.8p2
