A proof in Set theory about the number of different subset in a Set

[Shing]

hi guys, I am a physics major, recently doing a few self-reading on analysis, however, I got stuck at some excises of proofs.

Homework Statement

If a set A contains n elements, prove that the number of different subsets of A is equal to 2^n.

The Attempt at a Solution

First, I teared apart the problem, attempting to draw a general equation from looking at the numbers of different subset one by one.

E(m) regarded as the possible number of different subsets which contains m elements, whereas A has total n elements.

E(0)=1
E(1)=n
E(2)=\frac{n^2}{2}-\frac{n}{2} (this is obtained from a bit of analysis. Possibly be fault however.)
E(3)=\frac{5n^3}{6} -\frac{n^2}{2}-\frac{n}{3} (so is it.)
.
.
.
E(n)=1

However, I was almost driven crazy... I could't make the generl equation from it (which is hopefully 2^n)

Any other approaches? Or any mistakes I made during my approach?

Thanks for reading!

 

[Mark44]

hi guys, I am a physics major, recently doing a few self-reading on analysis, however, I got stuck at some excises of proofs.

Homework Statement

If a set A contains n elements, prove that the number of different subsets of A is equal to 2^n.

The Attempt at a Solution

First, I teared apart the problem, attempting to draw a general equation from looking at the numbers of different subset one by one.

E(m) regarded as the possible number of different subsets which contains m elements, whereas A has total n elements.

E(0)=1
E(1)=n
E(2)=\frac{n^2}{2}-\frac{n}{2} (this is obtained from a bit of analysis. Possibly be fault however.)
E(3)=\frac{5n^3}{6} -\frac{n^2}{2}-\frac{n}{3} (so is it.)

The formula above is incorrect. It should be (1/6)(n³ - 3n² + 6n)

.
.
.
E(n)=1

However, I was almost driven crazy... I could't make the generl equation from it (which is hopefully 2^n)

Any other approaches? Or any mistakes I made during my approach?

Thanks for reading!

In general, and using your notation, E(m) = the number of combinations of n things taken m at a time, or nCm, which is defined to be n!/(m! (n - m)!).

For example, if n = 3, there are:
1 set with nothing in it -- {}
3 sets with 1 element -- {1}, {2}, {3}
3 sets with 2 elements -- {1, 2}, {1, 3}, {2, 3}
1 set with 3 elements -- {1, 2, 3}

All together there are 1 + 3 + 3 + 1 subsets, or 8 subsets, or 2³ subsets.

By the same reasoning, it's easy to show that for a set with 4 elements, there are 2⁴ subsets. You can use mathematical induction to show what you're trying to show; namely that for a set of n elements, there are 2ⁿ subsets.

 

[Shing]

Thanks for answering!
However, I would love to know how to prove nCm suitable here?

 

[Hurkyl]

For the record, a direct counting argument is possible without any induction or summation identities -- you just need to find a different way to describe a subset than by first listing its size (say, m), and then by listing m elements.

 

[Shing]

Probably my knowledge in Math induction is not enough.
I have attempted to prove it via M.I.
however, I was having some hard time

my approach:
1.) obviously it is true for n=1,0
2.) I assume that it is true for 2^k, where k belongs to natural set.
3.) but how do I know/prove it is also true for 2^{k+1}? since I can't write such a function at all!

 

[Shing]

besides, I always do not understand how the nCm stuffs have to do with the so-called "possibilities" of combinations?
thanks!

 

[HallsofIvy]

Suppose A contains k+ 1 elements, with k> 0. Let a_0 be one of the elements of A. Every subset of A either contains a_0 or it does not. If it does not, it is a subset of A-\{x_0\} which has k elements so there are 2^k of them. If it does not, it is the same as a subset of A- \{x_0\} union x_0 so there are 2^k of them. Together there are 2^k+ 2^k= 2(2^k)= 2^{k+1} subsets of A.