Proving Symmetric Difference of A and B

Question

Let A and B be sets. Define the symmetric difference of A and B as A∆B= (A ∪ B) − (A ∩ B). (a) Prove that A∆B = (A − B) ∪ (B − A)

I tried to start this but am getting really lost. if someone could try to help that would be great

Answer 1

Hint: If $x \in A\Delta B$, then x is an element that’s in A or B, but not both. Compare this to $(A - B)\cup(B-A)$. Would $x$ be in this set as well?

Answer 2

The most direct way (not always the simplest) to prove that "$X= Y$", for sets, is to prove both $X\subseteq Y$ and $Y\subseteq X$. And to prove "$X\subseteq Y$" start with "if $x\in X$" then use the properties of sets X and Y to conclude "then $x\in Y$".

Here the two sets are $X= (A\cup B)- (A\cap B)$ and $Y= (A- B)\cup (B- A)$. If $x\in (A\cup B)- (A\cap B)$ then $x\in A\cup B)$ and $x\notin A\cap B$. That means x is in either A or B but not both.
Case 1: Suppose x is in A but not in B. Then x is in $A- B$ so in $(A- B)\cup (B- A)$. Case 2: Suppose x is in B but not in A. Then x is in $B- A$ so in $(A- B)\cup (B- A)$. In either case, $(A\cup B)- (A\cap B)\subseteq (A- B)\cup (B- A)$.

Now the other way- Suppose $x\in (A- B)\cup (B- A)$. Then either Case 1: $x\in A- B$. Then $x\in A\cup B$ but not in $A\cap B$ so $x\in (A\cup B)- (A\cap B)$. Case 2: $x\in B= A$. Then, again, $x\in A\cup B$ but not in $A\cap B$ so $x\in (A\cup B)- (A\cap B)$. In either case, $\cup (B- A)\subseteq (A\cup B)- (A\cap B)$.

Therefore, $(A\cup B)- (A\cap B)= (A- B)\cup (B- A)$.

Answer 3

If you are new to elementary set-theory the best way to prove this was given by user247327. However, if not then normally you should not make this any more difficult than it is.

Let $ A∆B=(A \cup B) \setminus (B \cap A) $. Remember DeMorgan: $A\setminus(B \cup C)= (A\setminus B) \cap (A\setminus C) $

We come down to :

$$ (A\cup B)\setminus (B \cap A)=((A\cup B)\setminus B)\cup ((A\cup B)\setminus A) $$
If you see what $(A\cup B)\setminus B$ and $(A\cup B)\setminus A$ is you are done.

Answer 4

Using that $A - B = A \cap B^C$:

$A \Delta B = $

$(A \cup B) - (A \cap B) = $

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

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

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

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

$(A - B) \cup (B - A)$