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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