What can be understood by $\rm A-B = B - A$ in set theory?
What does this tell us about the characteristics of $A$ and $B$ and their relationship? I'm quite confused as I am new to set theory.
$\endgroup$7 Answers
$\begingroup$Remember that "$A-B$" means "The set of things in $A$ but not in $B$." So "$A-B=B-A$" means "The things in $A$ but not $B$ are exactly the things in $B$ but not $A$."
Now, for which $A$ and $B$ is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation $A-B=B-A$ hold for $A=\{1, 2, 3\}=B$? What about $A=\{1,2,3\}, B=\{1, 2\}$? What about $A=\{1, 2, 3\},B=\{2, 3, 4\}$?
Based on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like "Assume $x\in A-B$. Then [stuff]. So $x\in B-A$. etc.")
$\endgroup$ 0 $\begingroup$You can work it out using a Venn diagram.
The above picture (sourced from Wikipedia) shows that $A \cup B$ can be divided into three regions: $A - B$ (the bit purely in orange), $A \cap B$ (the bit that's orange and blue), $B - A$ (the bit that's purely in blue).
What your equation is saying is that the bit purely in orange is equal to the bit purely in blue. Since the remaining bit in orange overlaps with the thing in blue, and the remaining bit in blue overlaps with the thing in orange, these two sets must be equal.
More formally, we see that $$A = (A - B) \cup (A \cap B)$$ $$B = (B - A) \cup (A \cap B)$$
Since $A - B = B - A$, we have that $$\begin{align} A &= (A - B) \cup (A \cap B)\\ &= (B - A) \cup (A \cap B)\\ &= B\end{align}$$
In fact, the equation $A - B = B - A$ is equivalent to the one $A = B$. Akiva explains below. $$ $$
$\endgroup$ 4 $\begingroup$($A-B=B-A$) means that the set of everything in $A$ which is not in $B$ equals the set of everything in $B$ which is not in $A$.
This is possible only when $\underline{(\phantom{A=B})}$ because:
For any element $x$ of $A-B$, we have $x\in A$ and $x\notin B$.
For any element $x$ of $B-A$, we have $x\in B$ and $x\notin A$.
But there is no element that can be in $A$ and not in $A$, and in $B$ and not in $B$.
Therefore $A-B$ is $\underline{\phantom{\quad\emptyset\quad}}$ , as is $B-A$. Meaning...
$\endgroup$ 0 $\begingroup$Two sets $X,Y$ are equal if and only if for all elements $z$ it holds that $z\in X\iff z\in Y$. If $z\in A-B$ then $z\in A$ but $z\notin B$ and if $z\in B-A$ then $z\in B$ but $z\notin A$. It seems impossible that an element could belongs to both $A-B$ and $B-A$, doesn't it?
$\endgroup$ $\begingroup$$A - B$ can be rewritten as $A - (A \cap B)$.
$B - A$ can be rewritten as $B - (A \cap B)$.
Given: $A - (A \cap B) = B - (A \cap B)$.
which means: $A$ and $B$ should be equal and $A - B = B - A = \emptyset$
$\endgroup$ 4 $\begingroup$OP says
I'm quite confused as I am new to set theory.
So wordy exposition really can't be harmful, especially since the OP asked "what can be understood", leaving the door open for a lengthy response.
If $a$ and $b$ are numbers, then $a - b = a$ is only possible if $b = 0$.
If $A$ and $B$ are sets, then $A - B = A$ if $B = \emptyset$, but that is not the end of the story. We know that $A - B \subseteq A$, and after some thought we come up with
Definition: An element $\hat a \in A$ is said to be safe from $B$ if $\hat a \in A-B$.
Exercise 1: $\hat a \in A$ is safe from $B \Leftrightarrow \hat a \notin B$.
Exercise 2: $A - B = A \Longleftrightarrow $ all elements in $A$ are safe from $B$.
Exercise 3: $A - B = A \Leftarrow \Rightarrow A \cap B = \emptyset$.
Exercise 4: $B - A = B \Leftarrow \Rightarrow A \cap B = \emptyset$.
Exercise 5: $A - B = A \Leftarrow \Rightarrow $ all elements in $B$ are safe from $A$.
So, we can summarize as follows,
Proposition 1: $A - B = A \Leftarrow \Rightarrow B - A = B \Leftarrow \Rightarrow A \text{ and } B\;$ are disjoint sets.
We see that there is certainly some symmetry going on here.
If $a$ and $b$ are numbers, then $a - b = b - a \Leftarrow \Rightarrow a = b$. Does this at least now carry over into set theory? What could go wrong with so much symmetry?
Proposition 2: Suppose for two sets $A$ and $B$,
$\tag 1 A - B = B - A$
Then $A = B$ and $A - B = B - A = \emptyset$.
Proof
Set $C = A - B = B - A$ and let $c \in C$. Then since $C = B - A$, $\,c \in B$. But we also have, $c \in A - B$, i.e. $c \in A$ is safe from $B$. By exercise 1, $c \notin B$, a contradiction.
So $C = \emptyset$. So, $A - B = \emptyset$. But this means no element in $A$ is safe from $B$, or, by exercise 1, every element in $A$ must be be in $B$, Same argument shows that every element in $B$ must be in $A$, so $A = B$. $\qquad \blacksquare$
All this symmetry here is really exciting, and before you know it you'll really appreciate the following material.
Definition: The symmetric difference ${\displaystyle A\,\triangle \,B}$ of two sets $A$ and $B$ is given by
$\tag 2 {\displaystyle A\,\triangle \,B=(A - B)\cup (B - A)}$
Exercise 6: Show that $A\,\triangle \,B=(A\cup B) - (A\cap B)$.
Exercise 7: Show that the following equalities are all equivalent statements,
$\qquad {\displaystyle A\,\triangle \,B = \emptyset}$
$\qquad A = B$
$\qquad A - B = B - A$
Note: The definition "$\hat a \in A$ is said to be safe from $B$" was made to aid intuition and you won't find it by googling. But see symmetric difference.
$\endgroup$ $\begingroup$I'll give you a simple explanation.
What if I told you that the answer is that $A - B = B - A = \emptyset$?
Assume that there is some element $x \in A - B \land x \in B - A$.
However, this means that $x \in A \cap B$. This is a contradiction. Therefore, it has to be the case that $A - B = B - A = \emptyset$.
I wish I was good at making animations, but you can imagine the intersection of the two sets as a black hole of sorts. Whenever, we place an element that exists in both differences, it just goes to the intersection. This proof is similar to the proof for showing that two equivalence classes are disjoint if you're interested.
$\endgroup$ 2
