# Consider a Set

Consider this set: *A* = {*a*, *b*, *c*, *d*}.
As defined in the first reading assignment, a subset of *A* is another set that contains only elements from the set *A*, but many not contain all the elements of *A*. A **proper subset** is a subset that is not identical to the original set—it contains fewer elements.
We can list all of the subsets of *A*:

{} (or Ø), {*a*}, {*b*}, {*c*}, {*d*}, {*a*,*b*}, {*a*,*c*}, {*a*,*d*}, {*b*,*c*}, {*b*,*d*}, {*c*,*d*}, {*a*,*b*,*c*}, {*a*,*b*,*d*}, {*a*,*c*,*d*}, {*b*,*c*,*d*}, {*a*,*b*,*c*,*d*}

*a*, either it’s in the set or it’s not. Thus there are 2 choices for that first element. Similarly, there are two choices for

*b*—either it’s in the set or it’s not. Using just those two elements, we see that the subsets are as follows:

{}—both elements are not in the set
{*a*}—*a* is in; *b* is not in the set
{*b*}—*a* is not in the set; *b* is in
{*a*,*b*}—*a* is in; *b* is in

*a*times the two for

*b*gives us 2

^{2}= 4 subsets. You can draw a tree diagram to see this as well. Now let’s include

*c*. Again, either

*c*is included or it isn’t, which gives us two choices. The outcomes are {}, {

*a*}, {

*b*}, {

*c*}, {

*a*,

*b*}, {

*a*,

*c*}, {

*b*,

*c*}, {

*a*,

*b*,

*c*}. Note that there are 2

^{3}= 8 subsets. Including all four elements, there are 2

^{4}= 16 subsets. 15 of those subsets are proper, 1 subset, namely {

*a*,

*b*,

*c*,

*d*}, is not. In general, if you have

*n*elements in your set, then there are 2

^{n}subsets and 2

^{n}− 1 proper subsets.

## Licenses & Attributions

### CC licensed content, Original

- Mathematics for the Liberal Arts I.
**Provided by:**Extended Learning Institute of Northern Virginia Community College**Located at:**https://online.nvcc.edu/.**License:**CC BY: Attribution.