5
$\begingroup$

Hello I am taking a Discrete Mathematics course and having some trouble with this question about sets:

Under what conditions is $A- B = B- A$

Diagram: $A- B$

Diagram: $B- A$

Maybe I'm understanding incorrectly, but how can A-B and B-A be equal if they contain elements that are not in each other?

  • 0
    You're exactly right: **if** they contain elements that are not in each other, they won't be equal :) – 2017-01-28
  • 0
    And if $A-B$ and $B-A$ cannot be equal if one contains elements not in the other, what cases are left? – 2017-01-28
  • 1
    JFYI: $B\backslash A$ is a more common notation for $B-A$, in case you ever encounter it. – 2017-01-28
  • 3
    @TheCount `\setminus` is a more common command to use; it makes the spacing look nicer: $B\setminus A$. Also, using $B-A$ is common enough, I'd say. It's important to be aware of both notations. – 2017-01-28
  • 3
    I disagree with the edit that changed the $A-B$ notation to $A \setminus B$, etc. As much as I dislike the $A-B$ notation, it's perfectly valid and need not be changed. And as Arthur said it's important to be aware of both. – 2017-01-28
  • 0
    @Arthur well that's what I meant, really, was to make the OP aware of the other standard. Thanks for the tip, BTW. – 2017-01-28
  • 0
    @tilper agreed. – 2017-01-28
  • 0
    @tilper Specifically, writing $B-A$ is not _wrong_ on a site like this, where people gather from all sorts of backgrounds. In a specific class the teacher can decide that one of them is incorrect, but that's not appropriate here. – 2017-01-28
  • 0
    No problem, again appears $-$ instead of $\setminus$ also "When" instead of "when". – 2017-01-28
  • 0
    See also: [For what conditions on sets $A$ and $B$ the statement $A - B = B - A$ holds?](https://math.stackexchange.com/q/1909611) – 2017-06-03
  • 0
    Presumably, $A-B$ is the relative complement, which might be more easily read if it were $A\setminus B$. Some clarification would be appreciated. – 2017-09-17
  • 0
    It's about sets – 2017-09-17

4 Answers 4

4

If $\color{blue}{A-B=B-A}$

$A\subset B$: $$A\subset (A-B)\cup B=(B-A)\cup B=(B\cap A')\cup B=B$$ similar way shows $B\subset A$ so $\color{blue}{A=B}$.

contrary is trivial.

  • 0
    It would have been helpful to say that A and B are **sets** in the first post! – 2017-01-28
  • 0
    Change what? No, I didn't change anything. – 2017-02-02
2

If $x\in A-B$, then $x\in A$ but $x\not\in B$. Similarly, if $x\in B-A$, then $x\in B$ but $x\not\in A$; so if $x\in A-B$ and $x\in B-A$, we conclude that $x\in A$ and $x\not\in A$, a contradiction! Hence there exists no such $x$, so $A-B=\emptyset$, so $A\subseteq B$, and similarly $B-A=\emptyset$, so $B\subseteq A$, and therefore $A=B$.

1

Suppose $B - A = A -B$.

Claim: $A = B$.

If not, this means there either an $x \in A$ such that $x \notin B$ (and then $x \in A - B$, but $x \notin B - A$, so $A- B \neq B-A$, contradiction), or there is an $x \in B$ with $x \notin A$ (and then $x \in B - A$, while $x \notin A - B$, so $A - B \neq B- A$, contradiction again).

So $A = B$ and $A - B = B- A = \emptyset$.

0

Let $a\in A$.

Consider two cases.

  1. $a\in A\setminus B$.

Hence, since $A\setminus B=B\setminus A$, we obtain $a\in B$.

  1. $a\not\in A\setminus B$.

Thus, $a\not\in A\cap \bar{B}$ or $a\in\bar{A}\cup B$, which says $a\in B$ again.

By the same way we can prove that if $a\in B$ then $a\in A$.

Id est, $A=B$.

Done!