Part A : Given |A\B| = |B\A|, prove |A|=|B|
Part B : Provide 2 sets (A and B) such that |A| = |B| but |A\B| is not the same as |B\A|
Part A:
Let A,B be sets. Given that |A\B| = |B\A| then |A|=|B|
Hint was given to us "find a one to one and onto function between A and B". but am struggling to do so."
I started with a bidirectional inclusion, i.e first show |A| <= |B| then vis-versa, as we only care about the cardinality and not the elements inside, I wasn't sure how to proceed. i.e let A have n elements, since |A\B| = |B\A|, B also needs to have at least n elements...
Part B:
For the 2nd part, after testing a lot of examples I came up with this: Let A = $\mathbb{Z}$, B = $\mathbb{N}$
|A| = |B| = $\aleph_0$
But A\B = Only the negative integers, |A\B| = $\aleph_0$
B\A = $\emptyset$, |B\A| = 0
Which clearly have different cardinalities
Is my logic flawed, and are there other examples (not so similar)?
Part A is true ONLY when $A$ & $B$ are finite.
In the Pictorial Proof ( Venn Diagram ) , where I am using numbers $x,y,z$ to indicate Set Sizes ( not elements ) , we can easily see that :
$|A|=x+y$
$|B|=z+y$
We are given $|A \setminus B|=x=z=|B \setminus A|$
Hence $x+y=z+y$ ( adding $y$ )
$|A|=x+y=z+y=|B|$
This type of calculation works when we have finite numbers.
Part B will require $A$ & $B$ infinite.
Let $A$ be some arbitrary infinite Set ( Eg Integers , rationals , irrationals , Naturals )
Let $C$ be some finite Set of unrelated elements , eg $C=\{ Apple , Orange , Newton , Atom \}$
Let $B=A \cup C$
The Sets $A$ & $B$ will work.
$|A|=|B|$
$|A \setminus B|=0$
$|B \setminus A|=|C|$