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 $\{ 2, 4, 6, 8, ... \} \text{ or } \{ 10, 20, 30, ... \}$

https://tenor.com/view/how-interesting-coffee-drin-drinking-nod-gif-4574690

 How clear is this post?