Proving Set Theory: A U B=(A-B) U (B-A)U (A∩B) | Homework Help"

BSo if x is in A then it must be in B. This is what you need to show. So let x be in A. You need to show that x is in B. So you need to show that x is in (A-B) \cup (B-A) \cup (A \cap B). So you will need to use the definition of (A-B), (B-A), and (A\cap B) to use the definition of the union.
  • #1
bonfire09
249
0

Homework Statement


Let A and B be sets. Prove that A U B=(A-B) U (B-A)U (A∩B)

Homework Equations


The Attempt at a Solution


If I want to prove the foward direction: A U B⊆(A-B) U (B-A)U (A∩B) then
from my understanding I know that xεA or xεB. And I can assume wolog that xεA. But since I assume that x is an element from A, its already equal to (A-B) U (B-A)U (A∩B)? Since xεA and x is not an element of B or xεb and not an element of A or x is both an element of A and B? is that all i have to show for the foward direction?
 
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  • #2
ok i was thinking this for a while. Let's take two sets A={1,2,3,4,5} and {3,4,5,6,7}
so I can divide this proof into 3 cases:
case 1: xεA
case 2: xεB
case 3: xεA and xεB.

am i right?
 
  • #3
bonfire09 said:
ok i was thinking this for a while. Let's take two sets A={1,2,3,4,5} and {3,4,5,6,7}
so I can divide this proof into 3 cases:
case 1: xεA
case 2: xεB
case 3: xεA and xεB.

am i right?

Im not sure what you've covered so far. Do you know the logical definition of a subset?
 
  • #4
Foward Direction:A U B⊆(A-B) U (B-A)U (A∩B)
Case 1: xεA
Since xεA then xε(A-B). xε(A-B) means that xεA and x is not an element of B.
Case 2: xεB
Since xεB then xε(B-A) which means that xεB and x is not an element of A.
Case 3:xεA and xεB
Since xεA and xεB then xε(A∩B) means that xεA and xεB.
Hence A U B⊆(A-B) U (B-A)U (A∩B).

Reverse direction
(A-B) U (B-A) U (A∩B)⊆A U B
Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

Case 1:xε(A-B)
Since xε(A-B) then this means that xεA and x is not a element of B. Which means that xε(AUB). xε(AUB) means that xεA or xεB. Hence (A-B)⊆A U B.

Case 2: xε(B-A)
Since xε(A-B) then this means that xεB and x is not an element of A. Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

Case 3: xε(A∩B)
Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB. Hence (A∩B)⊆A U B.
Therefore (A-B) U (B-A) U (A∩B)⊆A U B
Im not entirely sure if this proof is right. It seems verbose but then again I am taking an introduction class.
 
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  • #5
Not quite. Let me go through what you did.


Case 1: xεA

"Since xεA then xε(A-B)" (This is not true consider x in both A and B. Then x won't be in A-B but it is in A). xε(A-B) means that xεA and x is not an element of B.
Case 2: xεB
"Since xεB then xε(B-A)" (Again this is not true) which means that xεB and x is not an element of A.
Case 3:xεA and xεB
Since xεA and xεB then xε(A∩B) "means that xεA and xεB (redundant)".
Hence A U B⊆(A-B) U (B-A)U (A∩B).

Reverse direction
(A-B) U (B-A) U (A∩B)⊆A U B
Background: I have to show that xε(A-B) or xε(B-A) or xε(A∩B) is a subset of A U B.

Case 1:xε(A-B)
Since xε(A-B) then this means that xεA "and x is not a element of B (useless info)". Which means that xε(AUB). xε(AUB) "means that xεA or xεB (This step should be before you say x is in AUB). Hence (A-B)⊆A U B.

Case 2: xε(B-A)
Since xε(A-B) then this means that xεB "and x is not an element of A (useless info)". Hence xε(AUB) which is similar as the above case. Hence (B-A)⊆A U B.

Case 3: xε(A∩B)
Since xε(A∩B) which implies that xεA and xεB. This means that xε(AUB). Since xεA or xεB or both xεA and xεB (this step should be before concluding xε(AUB)) . Hence (A∩B)⊆A U B.
Therefore (A-B) U (B-A) U (A∩B)⊆A U B
Im not entirely sure if this proof is right. It seems verbose but then again I am taking an

The first direction needs more work. The second just needs a little fix. The first direction is harder and it might require showing that A = (A-B)U(A∩B). Also, I'm not sure if your class covered this but, xε(A-B) <=> xε(A∩B[itex]^{c}[/itex]) (B[itex]^{c}[/itex]) is the complement of B)
 
  • #6
since the definition of AUB is xεA or xεB then I thought that x could be in either set A or set B or both. That was the reason behind my three cases. Oh maybe I should have written it like this for the foward direction

Case 1: xεA and x is not element of B
Case 2: xεA and xεB
Case 3: xεB and x is not an element of A but that is the same as case 1.
 
  • #7
bonfire09 said:
since the definition of AUB is xεA or xεB then I thought that x could be in either set A or set B or both. That was the reason behind my three cases. Oh maybe I should have written it like this for the foward direction

Case 1: xεA and x is not element of B
Case 2: xεA and xεB
Case 3: xεB and x is not an element of A but that is the same as case 1.

I'm not sure what you are trying to justify.
Here is the definition of a subset:
A[itex]\subset[/itex]B [itex]\Leftrightarrow[/itex] for any x, x [itex]\in[/itex]A [itex]\Rightarrow [/itex] x [itex]\in[/itex]B
 

1. What is a set theory proof?

A set theory proof is a logical argument that uses the axioms and rules of set theory to show that a statement is true. It involves breaking down the statement into smaller, more manageable parts and using the properties of sets to demonstrate its validity.

2. How do I know if my set theory proof is correct?

A set theory proof is considered correct if it follows the rules of logic and the axioms of set theory. This can be verified by checking each step of the proof to ensure that it logically follows from the previous steps and that it uses valid set theory principles.

3. What are some common strategies for solving set theory proofs?

Some common strategies for solving set theory proofs include using Venn diagrams, breaking down the statement into smaller subsets, and using properties of specific sets (such as the empty set or universal set).

4. How do I know which axioms and rules to use in a set theory proof?

In general, the axioms and rules used in a set theory proof will depend on the specific statement being proved. It is important to carefully read and understand the statement before beginning the proof, and to have a solid understanding of the axioms and rules of set theory.

5. Can I use set theory to prove any mathematical statement?

No, set theory is not applicable to all mathematical statements. It is primarily used for proving statements related to sets, such as set equality, subset relationships, and set operations. Other branches of mathematics, such as calculus or geometry, have their own specific methods of proof.

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