Prove: Sets Union & Intersection Hypothesis

[VinnyCee]

DISCRETE MATH: Prove a "simple" hypothesis involving sets. Use mathematical induction

Homework Statement

Prove that if A_1,\,A_2,\,\dots,\,A_n and B are sets, then
\left(A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_n\right)\,\cup\,B\,=\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_n\,\cup\,B\right)

Homework Equations

A\,\cap\,B\,=\,B\,\cap\,A <----- commutative law

A\,\cup\,\left(B\,\cap\,C\right)\,=\,\left(A\,\cup\,B\right)\,\cap\,\left(A\,\cup\,C\right) <----- distributive law

The Attempt at a Solution

I don't know how to start this other than that I need to use the two laws above. Maybe change the notation? I don't know.

\bigcap_{i\,=\,1}^{n}\,A_i\,\cup\,B\,=\,\left(A_1\,\cap\,B\right)\,\cup\,\left(A_2\,\cap\,B\right)\,\cup\dots\,\cup\,\left(A_n\,\cap\,B\right)

What should be the next step or is there a better way of going about this?NOTE: For LaTeXers, \cup is a union and \cap is an intersection.

 

[mjsd]

your title says use mathematical induction... so did u try using that method to do this question?

 

[VinnyCee]

Let P(n) be \left(A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_n\right)\,\cup\,B\,=\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_n\,\cup\,B\right)

Then \forall\,n\,\left(P(n)\right), right?
First I do the basis step for P(1):

A_1\,\cup\,B\,=\,\left(A_1\,\cup\,B\right)
Now I need to show that P(k)\,\longrightarrow\,P(k\,+\,1)?

For P(k) (Also, assume it is true for induction):

\left(A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_k\right )\,\cup\,B\,=\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A _k\,\cup\,B\right)EDIT: Removed errors

 

[matt grime]

1. Don't start by writing out what you want to prove (just after the 'For P(k+1)').

2. P(k) is a statement that two things are equal. So why have you used the symbol P(k) as if it were a set?

3. Use strong induction.

4. If that doesn't mean anything to you then consider this:

take AnBnCnDnEnF. Let X=AnBnCnDnE and Y=F. Then we can rewrite that as XnY. I've gone from a statement about 6 sets to one about 2 sets. Now can I use anything I know to be true for 2 sets...

 

[VinnyCee]

Strong induction is covered in the next section. It says that it is easier to use strong induction in many cases, I am sure this is one, but we have to use "mathematical induction". Isn't strong induction just a special case of mathematical induction?

Let S be A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_k

\left(A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_k\,\cap \,A_{k\,+\,1}\right)\,\cup\,B\,=\,\left(S\,\cap\,A_{k\,+\,1}\right)\,\cup\,B

Then use commutative law:

B\,\cup\,\left(S\,\cap\,A_{k\,+\,1}\right)

Now use distributive law:

\left(B\,\cup\,S\right)\,\cap\,\left(B\,\cup\,A_{k\,+\,1}\right) < ----- Equation 1

Now, assuming the p(k) induction:

\left(B\,\cup\,S\right)\,\equiv\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_k\,\cup\,B\right)

Substitute this assumption into equation 1 above:

\left(B\,\cup\,S\right)\,\cap\,\left(B\,\cup\,A_{k\,+\,1}\right)\,\equiv\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_k\,\cup\,B\right)\,\cap\,\left(B\,\cup\,A_{k\,+\,1}\right)

Now use the commutative law again on the last two terms:

\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_k\,\cup\,B\right)\,\cap\,\left(A_{k\,+\,1}\,\cup\,B\right)

And this proves that for P(k\,+\,1):

\left(A_1\,\cap\,A_2\,\cap\,\dots\,\cap\,A_k\,\cap\,A_{k\,+\,1}\right)\,\cup\,B\,=\,\left(A_1\,\cup\,B\left)\,\cap\,\left(A_2\,\cup\,B\right)\,\cap\,\dots\,\cap\,\left(A_k\,\cup\,B\right)\,\cap\,\left(A_{k\,+\,1}\,\cup\,B\right)

Is the proof correct?