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For a set \(S\), the power set, which is written as \(P(S)\) or \(2^S\) refers to the collection of all subsets of \(S\).
For example, \(2 ^ {\lbrace 10, 20 \rbrace} = \lbrace \lbrace \rbrace, \lbrace 10 \rbrace, \lbrace 20 \rbrace, \lbrace 10,20 \rbrace \rbrace \).
How big is \(2^S\)? Well, if \(S\) has \(n\) elements, the power set \(2^S\) has \(2^n\) elements:
\[|2^S| = 2^{|S|}\]
Hence the notation.
Comments
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 \(2^n\).
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\\).Another proof is by induction on n.
The base case is for n = 0. That means S = {}. So the claim to be shown is:
\[|2^{\lbrace \rbrace}| = 2^{|\lbrace \rbrace|} \]
\(2^{\lbrace \rbrace}\) is all subsets of \(\lbrace \rbrace\), which just contains one member \(\lbrace \rbrace\). So the left side equals 1. On the right, since \(|\lbrace \rbrace|\) equals 0, we also get 1.
For the induction step, we'll show that increasing \(S\) by one element doubles the size of \(2^S\).
Let \(S' = S \cup \lbrace x \rbrace\). To each subset \(A\) of \(S\) we can associate a pair of subsets of \(S'\) -- \(A\) itself, and \(A \cup \lbrace x \rbrace\). And this mapping is unique, in that every subset of \(A'\) gets hit by the mapping, and gets hit only once. So the number of subsets of \(A'\) is exactly twice the number of subsets of \(A\).
Another proof is by induction on n. The base case is for n = 0. That means S = {}. So the claim to be shown is: \\[|2^{\lbrace \rbrace}| = 2^{|\lbrace \rbrace|} \\] \\(2^{\lbrace \rbrace\}\\) is all subsets of \\(\lbrace \rbrace\\), which just contains one member \\(\lbrace \rbrace\\). So the left side equals 1. On the right, since \\(|\lbrace \rbrace|\\) equals 0, we also get 1. For the induction step, we'll show that increasing \\(S\\) by one element doubles the size of \\(2^S\\). Let \\(S' = S \cup \lbrace x \rbrace\\). To each subset \\(A\\) of \\(S\\) we can associate a pair of subsets of \\(S'\\) -- \\(A\\) itself, and \\(A \cup \lbrace x \rbrace\\). And this mapping is _unique_, in that every subset of \\(A'\\) gets hit by the mapping, and gets hit only once. So the number of subsets of \\(A'\\) is exactly twice the number of subsets of \\(A\\).The notation \(2^S\) also agrees with the exponential notation \(B^A\) for general sets \(A\), \(B\), which is defined as the set of all functions from A to B.
In this context, let's take 2 to mean the binary set \(\lbrace True,False \rbrace \).
So \(2^S\) means all functions going from \(S\) into \(\lbrace True,False\rbrace \). Each of these is a predicate, assigning \(True\) or \(False\) to every member of \(S\). And each predicate is equivalent to the subset of values in \(S\) which get mapped to \(True\).
So as a set of functions \(2^S\) is equivalent to the power set \(2^S\).
The notation \\(2^S\\) also agrees with the exponential notation \\(B^A\\) for general sets \\(A\\), \\(B\\), which is defined as the set of all functions from A to B. In this context, let's take 2 to mean the binary set \\(\lbrace True,False \rbrace \\). So \\(2^S\\) means all functions going from \\(S\\) into \\(\lbrace True,False\rbrace \\). Each of these is a predicate, assigning \\(True\\) or \\(False\\) to every member of \\(S\\). And each predicate is equivalent to the subset of values in \\(S\\) which get mapped to \\(True\\). So as a set of functions \\(2^S\\) is equivalent to the power set \\(2^S\\)..
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