# A Set Theory Proof

## Homework Statement

Homework Statement are located in the pdf below. I also upload the file onto an online viewer for pdf

Here is the link: http://view.samurajdata.se/psview.php?id=c1f5a372&page=1 [Broken]

None.

## The Attempt at a Solution

The attempt at a solution is located in the pdf below.

Here is the link again for those who do not wish to download the file: http://view.samurajdata.se/psview.php?id=c1f5a372&page=1 [Broken]

Can anyone see if I did anything wrong in the proof? I am not talking about the format of the proof, just the underlying logic. Thanks!

#### Attachments

• 76.4 KB Views: 68
Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
Mark44
Mentor
If you can use deMorgan's Laws, this can be done quite a bit shorter.

$$B - A = B \cap A^C$$

So
$$A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)$$

The latter expression is just U, the universal set.

If you can use deMorgan's Laws, this can be done quite a bit shorter.

$$B - A = B \cap A^C$$

So
$$A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)$$

The latter expression is just U, the universal set.
Yes, I know that. I should have mentioned it before hand that I wanted to use the "choose-an-element method" to prove it rather than algebraic manipulation. Other than that, do you think that there was anything wrong with my proof in the pdf?

Mark44
Mentor
IMO, you are overcomplicating things by using contradiction proofs. Why don't you just prove the two statements directly?

IMO, you are overcomplicating things by using contradiction proofs. Why don't you just prove the two statements directly?
To be honest, I do not really know how to prove one of the statements.

That is, A + B is a subset of A + (B - A).

The plus signs means "union."

Attempt:

1) Let x be in A + B.
2) We can say x is in A or x is in B.
3) If x is in A, then x is in A + (B - A).
4) If x is in B, then x is B and x can either be in A or not in A.
5) If x is in B and x is in A, then x is in A + (B - A).
6) If x is in B and x is not in A, then x is in (B - A).

I do not even know if I got it right...

Last edited:
Mark44
Mentor
To be honest, I do not really know how to prove one of the statements.

That is, A + B is a subset of A + (B - A).
How about writing the expression on the right as
$$B \cap A^C$$

The plus signs means "union."

Attempt:

1) Let x be in A + B.
2) We can say x is in A or x is in B.
x could be in both sets, if it is in their intersection.

Look at three different cases:
1) x is in A but not in B.
2) x is in B but not in A.
3) x is in $A \cap B$.

For each case, show that x is also in A U (B - A).
3) If x is in A, then x is in A + (B - A).
4) If x is in B, then x is B and x can either be in A or not in A.
5) If x is in B and x is in A, then x is in A + (B - A).
6) If x is in B and x is not in A, then x is in (B - A).

I do not even know if I got it right...