# Math Help - Set Theory Proof

1. ## Set Theory Proof

I am stuck on this problem, I am just simply awful with set theory proofs, I would really like anyone's help please

Let A,B and C be sets. Prove that (AUB) - C must be a subset of [A- (BUC)]U[B-(A intersect C)], but that equality need not hold.

Thanks!

2. Originally Posted by danio
I am stuck on this problem, I am just simply awful with set theory proofs, I would really like anyone's help please

Let A,B and C be sets. Prove that (AUB) - C must be a subset of [A- (BUC)]U[B-(A intersect C)], but that equality need not hold.

Thanks!
as always, to show that one set is a subset of another, you must show that every element in the first set is found in the other as well. thus, you must show,

$x \in (A \cup B) - C \implies x \in [A - (B \cup C)] \cup [B - (A \cap C)]$

try that and see where you get.

as for showing that equality need not hold, just come up with an example where you have the described subset relationship without equality. having trouble coming up with such an example, consider drawing Venn diagrams to come up with an idea. you will see what you need to do

3. Okay so I tried it and this is what I have so far:

Assume x is an element it [(AUB)-C]

Then x is not an element of C and x is an element of A and/or B.
Case 1: x is an element of A and not B
Then x is an element of [A-(BUC)]
Therefore x is an element [A-(BUC)]U[B-(A intersect C)]

Case 2: x is an element of B and not A
Then x is an element of [B-(A intersect C)]
Therefore x is an element of [A-(BUC)]U[B-(A intersect C)]

Case 3: x is an element of A and B
Then x is an element of [B-(A intersect C)]
Therefore x is an element of [A-(BUC)]U[B-(A intersect C)]

Therefore if x is an element of [(AUB) - C] then x is always an element of [A-(BUC)]U[B-(A intersect C)], hence [(AUB)-C] is a subset of [A-(BUC)]U[B-(A intersect C)].

Now I am not sure if this is correct bc I don't know if I am using the "-" part correctly. I assumed when you say A-B in sets it would be all the elements in A and not B but then I realized you would just say "\"

4. I also have this for the second part:

Equality need not hold for example when:
$A$ = {1,2,3}
$B$ = {4,5,6}
$C$ = {2,4}

Then $[(A \cap B)-C] = \{ 1,3,5,6 \}$ and $[A-(B \cup C)] \cup [B-(A \cap C)] = \{1,3,4,5,6 \}$. In which case it is a subset of but not equal too.

5. Originally Posted by danio
Okay so I tried it and this is what I have so far:

Assume x is an element it [(AUB)-C]

Then x is not an element of C and x is an element of A and/or B.
Case 1: x is an element of A and not B
Then x is an element of [A-(BUC)]
Therefore x is an element [A-(BUC)]U[B-(A intersect C)]

Case 2: x is an element of B and not A
Then x is an element of [B-(A intersect C)]
Therefore x is an element of [A-(BUC)]U[B-(A intersect C)]

Case 3: x is an element of A and B
Then x is an element of [B-(A intersect C)]
Therefore x is an element of [A-(BUC)]U[B-(A intersect C)]

Therefore if x is an element of [(AUB) - C] then x is always an element of [A-(BUC)]U[B-(A intersect C)], hence [(AUB)-C] is a subset of [A-(BUC)]U[B-(A intersect C)].

Now I am not sure if this is correct bc I don't know if I am using the "-" part correctly. I assumed when you say A-B in sets it would be all the elements in A and not B but then I realized you would just say "\"
yes, this works. this is not the way i did it, but my proof essentially went through the same steps. also, "-" is the same as "\", so you are good. A - B means the set of elements that are in A but not in B. A\B means the same thing. it is an alternate notation

Originally Posted by danio
I also have this for the second part:

Equality need not hold for example when:
$A$ = {1,2,3}
$B$ = {4,5,6}
$C$ = {2,4}

Then $[(A {\color{red}\cup} B)-C] = \{ 1,3,5,6 \}$ and $[A-(B \cup C)] \cup [B-(A \cap C)] = \{1,3,4,5,6 \}$. In which case it is a subset of but not equal too.
yes, except you should have union and not intersection where i put the red.