SUBSET

[Samin]

1.What is the number of all possible subsets of the set A={e,g,f}?

{e} {g} {f} {eg} {gf} {ef} {egf} φ so number of subsets = 8
Correct me if i’m wrong

 

[Harry_the_cat]

Yes there are 8 subsets.

 

[Samin]

Thank

Yes there are 8 subsets.

Thanks

 

[HallsofIvy]

In general if a set has n members then it has $\displaystyle 2^n$ subsets. $\displaystyle 2^3= 8$.

 

[Samin]

T

In general if a set has n members then it has $\displaystyle 2^n$ subsets. $\displaystyle 2^3= 8$.

This method looks easier
Thanks?

 

[HallsofIvy]

You can prove that by induction on the number of members.

Every set has at least the empty set and the entire set as subsets. If n= 0 the entire set is the empty set so the empty set has 1 subset. $\displaystyle 2^0= 1$.

If a set has exactly one member, {x}, then its subsets are precisely the empty set, { }, and the entire set, {x}, so has 2 subsets. $\displaystyle 2^1= 2$.

Suppose that any set with k members, k> 0, has $\displaystyle 2^k$ subsets. Let X be a set with k+ 1 members. Since k> 1, we can select a member of the set, x, and X- {x} is one of two kinds- either it contains x or it doesn't. All subsets that do not contain x are subsets of X-{x} and there are $\displaystyle 2^k$ of them. All subsets that do contain x are precisely a subset that does not contain x union {x} so there are also $\displaystyle 2^k$ of them. There are $\displaystyle 2^k+ 2^k= 2(2^k)= 2^{k+ 1}$.

For example, consider the set {a, b, c}. Select the member "c" and remove it. That leave {a, b} which has $\displaystyle 2^2= 4$ subsets: { }, {a}, {b}, and {a, b}. Those are all also subsets of {a, b, c}. Adding "c" to each of those gives {c}, {a, c}, {b, c}, and {a, b, c}, the other four subsets of {a, b, c} for a total of $\displaystyle 2^3= 8$ subsets.