Recall powers (or exponents) of numbers: 2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32

Similarly, sets have the power operation to create new sets.

def^n If A is a set, then the power set of A is another set denoted as

\mathbb{ P }(A) = \text{ set of all subsets of A } = \{ x: x \subseteq A \}

Recall: A is a subset of B if every element in A is also in B. Furthermore, if A is a finite set with n-elements, then we can find the number of subsets in A by using this formula:

2^n

To find the power set of A, we write a list of all the subsets of A first – remembering that:

  • the empty set is a subset of every set,
  • and every set is a subset of itself

Let’s look at some examples:

e.g.1. A = \{1, 2, 3 \}

Using the formula 2^n , we know that there are 2^3 = 8 possible subsets of A, namely:

\varnothing,  \{1, 2, 3 \},  \{1 \},  \{2 \},  \{3 \},  \{1, 2 \}, \{2, 3 \} \text{ and } \{1, 3 \}

Hence the power set is the set that contains all the above subsets:

\mathbb{ P }(A) = \{ \varnothing,  \{1, 2, 3 \},  \{1 \},  \{2 \},  \{3 \},  \{1, 2 \}, \{2, 3 \},  \{1, 3 \}  \}

Note: The cardinality (size) of  \mathbb{ P }(A)  = 8 = 2^3 where size of A= 3 elements

e.g.2.  B = \{1, 3 \}

We know there are 2^2 = 4 possible subsets of B, namely:

\varnothing,  \{1 \},  \{3 \} \text{ and }  \{1, 3 \}

Hence the power set is: \mathbb{ P }(B) = \{ \varnothing, \{1 \},  \{3 \} ,  \{1, 3 \}  \}

Note: Again, the cardinality of \mathbb{ P }(B) = 4 = 2^2 where set |B| = 2 elements

The cardinality of a power set A, where A is a finite set, is denoted as | \mathbb{ P }(A) |.

i.e. | \mathbb{ P }(A) |= 2^{|A|} \text{ where } | A | \text{ denotes the size ( or cardinality) of A }

e.g. Find the cardinality of the following power sets

  1. \mathbb{ P }(\{ 1\})

Set A = \{ 1\} in this case. We know we have 1 element in the set (i.e. |A| =1), hence 2^1 = 2  \text{ subsets and } | \mathbb{ P }( \{ 1\} ) | = 2^1  = 2 .

The subsets are \varnothing  \text{ and } \{ 1\}, hence the power set is \mathbb{ P }(\{ 1\}) = \{  \varnothing, \{ 1\} \}

2. \mathbb{ P }(\{ \varnothing \})

Set A = \{ \varnothing \} in this case hence, again, we have 1 element in the set (i.e. |A| =1). So 2^1 = 2   subsets and | \mathbb{ P }( \{ 1\} ) | = 2^1  = 2.

The subsets are \varnothing  \text{ and } \{ \varnothing\}, hence the power set is \mathbb{ P }(\{ \varnothing \}) = \{  \varnothing, \{ \varnothing \} \}

3. \mathbb{ P }( \varnothing).

Set A = \varnothing in this case, hence we have 0 elements in the set (i.e. |A| =0). So 2^0= 1  \text{ subsets and } | \mathbb{ P }( \varnothing ) | = 2^0  = 1 .

So the subset is only the empty set, { }, and the power set \mathbb{ P }( \varnothing) = \{ \varnothing \}

4. \mathbb{ P }( \{ \mathbb{ Z }, \mathbb{ N} \} ).

Set A = \{ \mathbb{ Z }, \mathbb{ N} \}  in this case. We have |A| = 2 = |\{ \mathbb{ Z }, \mathbb{ N } \}|, hence 2^2= 4 subsets and | \mathbb{ P }(  \{ \mathbb{ Z }, \mathbb{ N } \} ) | = 2^2  = 4 .

Hence the power set is \mathbb{ P }(\{ \mathbb{ Z }, \mathbb{ N} \}) = \{  \varnothing, \{ \mathbb{ Z }\},  \{ \mathbb{ N} \}, \{\mathbb{ Z }, \mathbb{ N} \} \}

If a set A is infinite, we cannot find all its subsets, hence cannot list its power set.

e.g. \mathbb{ P } ( \mathbb{ Z } ) instructs for a power set of an infinite set, the integers: \mathbb{ Z }. Hence we cannot find the power set. We can, however,  list some subsets:

  • the empty set, \varnothing
  • the set of integers \mathbb{ Z }
  • and any subset of integers, for instance \{ 2, 4, 6, 8, ... \} \text{ or } \{ 10, 20, 30, ... \}
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