• Support PF! Buy your school textbooks, materials and every day products Here!

A Set Theory Proof

  • Thread starter number0
  • Start date
  • #1
104
0

Homework Statement



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

for those people who are afraid to download attachments.

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


Homework Equations



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

Last edited by a moderator:

Answers and Replies

  • #2
33,485
5,174
If you can use deMorgan's Laws, this can be done quite a bit shorter.

[tex]B - A = B \cap A^C[/tex]

So
[tex]A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)[/tex]

The latter expression is just U, the universal set.
 
  • #3
104
0
If you can use deMorgan's Laws, this can be done quite a bit shorter.

[tex]B - A = B \cap A^C[/tex]

So
[tex]A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)[/tex]

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?
 
  • #4
33,485
5,174
IMO, you are overcomplicating things by using contradiction proofs. Why don't you just prove the two statements directly?
 
  • #5
104
0
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:
  • #6
33,485
5,174
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
[tex]B \cap A^C[/tex]

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 [itex]A \cap B[/itex].

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...
 

Related Threads on A Set Theory Proof

  • Last Post
Replies
20
Views
1K
  • Last Post
Replies
13
Views
5K
  • Last Post
2
Replies
25
Views
3K
  • Last Post
Replies
1
Views
948
Replies
2
Views
7K
Replies
13
Views
2K
Replies
5
Views
17K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
Top