For any sets $A$ and $B$ show that: $(A \cup B) \diagdown (A \cap B) = (A \diagdown B) \cup (B \diagdown A)$

Question

I have the question "solved" and I understand the what the question is asking I just feel like the solution I have is not sufficiently "mathematical" and would like some guidance writing it in a more sophisticated way. If the way I answered it is sufficient for full marks, however, that would be good to know as well. It just feels too... casual.

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The goal of this problem is to show that the set on the LHS of the equation is a subset of the set and the right and that the converse is true as well.

So the set on the LHS of the equation can be written in set builder notation as:

$(A \cup B) \diagdown (A \cap B)$ $=$ {$x: x\in A$ or $x\in B, x\notin A$ and $B$}

Similarly the set on the RHS of the equation can be written in set builder notation as

$(A \diagdown B) \cup (B \diagdown A)$ $=$ {$x: x\in A$, $x \notin B$ or $x \in B$, $x\notin A$}

If these set-builder notation write ups are wrong please let me know.

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1. To show LHS is a subset of RHS we let $x \in$ {$x: x\in A$ or $x\in B, x\notin A$ and $B$} and note that $x$ can be an element of either $A$ or $B$ but not both simultaneously, thus we can see that $x \in$ $(A \diagdown B) \cup (B \diagdown A)$

2. To show RHS is a subset of LHS we let $x \in$ {$x: x\in A$, $x \notin B$ or $x \in B$, $x\notin A$} and note that if $x \in A$ then $x$ is not an element of $B$ and if $x \in B$ then $x$ is not an element of $A$ hence $x$ could never be an element of $A$ and $B$ and thus $x \in$ $(A \cup B) \diagdown (A \cap B)$

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I feel like I've done nothing more than write the sets in set builder notation and then describe them and say this equals that. I can see how they are equivalent but it doesn't seem like enough work. Am I missing somethign here?

Any help is greatly appreciated.

Answer 1

According to the set operations' properties, we get that

\begin{align*} (A\cup B)\backslash(A\cap B) & = (A\cup B)\cap(A\cap B)^{c}\\\\ & = (A\cup B)\cap(A^{c} \cup B^{c})\\\\ & = ((A\cup B)\cap A^{c})\cup((A\cup B)\cap B^{c})\\\\ & = ((A\cap A^{c})\cup(B\cap A^{c}))\cup((A\cap B^{c})\cup(B\cap B^{c}))\\\\ & = (\varnothing\cup(B\cap A^{c}))\cup((A\cap B^{c})\cup\varnothing)\\\\ & = (B\cap A^{c})\cup(A\cap B^{c})\\\\ & = (A\cap B^{c})\cup(B\cap A^{c})\\\\ & = (A\backslash B)\cup(B\backslash A) \end{align*} and we are done.

Answer 2

One way without using complement. Definition for $(A \cup B) \setminus (A \cap B)$ gives $$(x\in A \lor x\in B)\land \neg (x\in A \lor x\in B) =(x\in A \lor x\in B)\land (x\notin A \land x\notin B)\quad (1) $$ and definition for $(A \setminus B) \cup (B \setminus A)$ is $$(x \in A \land x\notin B)\lor (x \in B \land x\notin A)\quad (2)$$ Let's start with $(1)$ and finish with $(2)$ : $$(x\in A \lor x\in B)\land (x\notin A \land x\notin B) = \\ =\Big(x\in A \land (x\notin A \land x\notin B) \Big)\lor \Big(x\in B \land (x\notin A \land x\notin B)\Big) =\\ =(x \in A \land x\notin B)\lor (x \in B \land x\notin A)$$

Answer 3

As long as the verbage is clear, its fine.   A proof works when it convinces the reader that each step is justified.

Personally, I discourage overly mixing math notations and words.   If you are going to use a verbose proof, just describe the sets through an arbitrary element. Something like:-

Take any element that is in either $A$ or $B$ but not in both simultaneously; thus we can see that this element is either in $A$ but not in $B$, OR in $B$ but not in $A$.

Therefore $(A\cup B)\smallsetminus(A\cap B)~\subseteq~(A\smallsetminus B)\cup(B\smallsetminus A)$.

Conversely...

Likewise, if you are going to use set notation, then do so an algebraic proof. (Hint: Distribution )

$${\quad(A\cup B)\smallsetminus(A\cap B)\\=\{x: (x\in A\lor x\in B)\land(x\notin A\lor x\notin B)\}\\~~\vdots\phantom{\tiny\{x: (x\in A\land x\notin A)\lor(x\in A\land x\notin B)\lor(x\in B\land x\notin A)\lor(x\in B\land x\notin B)\}}\\=\{x:(x\in A\land x\notin B)\lor(x\in B\land x\notin A)\}\\=(A\smallsetminus B)\cup(B\smallsetminus A)}$$

But that's just my preference.

Answer 4

$$A\Delta B =(A \cup B) \diagdown (A \cap B) = (A \diagdown B) \cup (B \diagdown A)$$ The set is called symmetric difference of $A$ and $B$

Note that on both sides we have those elements which are in one and only one of the two sets so they are equal.

The symmetric difference is also defined for three or more sets.

One notable property of the symmetric difference is the associative property $$ A\Delta (B\Delta C)=(A\Delta B)\Delta C$$ which makes a good exercise for students of a set theory course.