List the power set € P(S) for the set € S = {A,B,C,D,F}?
24.Describe
the
following
set
using
set
builder
notation:
A = {5,15,25,35,45,55,65}
I am not sure what the symbol "€" means. If a set has n elements, its powerset has 2ⁿ elements. So, P(S) has 32 elements including the empty set, 5 sets containing 1 element each, 10 elements containing 2 elements each, etc. Make sure you understand what a power set is.
This can be defined as a set of elements of the form 10n + 5 for integer n between 0 and 6.
As for listing all $\displaystyle 2^5= 32$ subsets, I recommend starting by listing the "singleton" sets:
{} (the empty set), {A},{B}, {C},{D}, {F},
now include each element of the original set "past" that one: {AB}, {AC}, {AD}, {AF}, {BC}, {BD}, {BF}, {CD}, {CF}, {DF}.
(The point of "past" is to avoid duplicates.)
Now, add a letter to each of those: {ABC}, {ABD}, {ABF}, {ACD}, {ACF}, {ADF}, {BCD}, {BCF}, {BDF}, {CDF}
etc.
As a check, the number of subsets containing n elements is $\displaystyle \begin{pmatrix}5 \\ n\end{pmatrix}$. So there is $\displaystyle \frac{5!}{0!5!}= 1$ subset with 0 elements, $\displaystyle \frac{5!}{1!4!}= 5$ subsets with 1 element, $\displaystyle \frac{5!}{2!3!}= 10$ subsets with two elements, $\displaystyle \frac{5!}{3!2!}= 10$ subsets with three elements,$\displaystyle \frac{5!}{4!1!}= 5$ subsets with 4 elements, and $\displaystyle \frac{5!}{5!0!}= 1$ subsets with 5 elements (the entire set).
For {5, 15, 25, 35, 45, 55, 65}, notice that 5= 1(5), 15= 3(5), 25= 5(5), 35= 7(5), 45= 9(5), 55= 11(5), 65= 13(5), odd multiples of 5. And those odd numbers can be written as 2n+1 for n from 0 to 6.