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So if I have set a = {1,2} and set b {2,1}

(a-b)-(b-a) What would be the resulting product?

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    $\begingroup$ The empty set. Order doesn't matter in sets, so a = b. a-b=b-a = a-a = b-b the empty set. So empty set - empty set = empty set. $\endgroup$ – fleablood Aug 5 '16 at 20:58
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If by $a-b$ you mean set $a$ without the elements in set $b$, then the answer if the empty set, because both sets have the same elements.

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  • $\begingroup$ Yeah but is 1,2 and 2,1 two distinct sets? such as a{{1,2}} and b{{2,1}} so they are two different sets. Thus set a={c} minus b={d} is still c. right? $\endgroup$ – JanCamila Aug 5 '16 at 20:39
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    $\begingroup$ Order of elements doesn't matter in sets. $\endgroup$ – bob.sacamento Aug 5 '16 at 20:39
  • $\begingroup$ No, they are not distinct sets. {1,2} and {2,1} are the exact same set. Order does not matter. $\endgroup$ – fleablood Aug 5 '16 at 20:59
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$a-b := \{x\in a \space AND \space x\notin b \} = \{\}$

$b-a := \{x\in b \space AND \space x\notin a \} = \{\}$

$\implies (a-b)-(b-a)=\{\}-\{\}=\{\}$

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In my living room I have three books. "Moby Dick", "Horton Hears a Who" and "Pink Honk-Honk". My roommate came around and shuffled them into a different order. Now my three books are "Horton Hears a Who", "Moby Dick", and "Pink Honk-Honk".

Has anything changed? I still have the same three books in my living room.

Let A = {the books in my living room} = {"Moby Dick", "Horton Hears a Who", "Pink Honk-Honk"}

Let B = {the books in my living room} = {"Horton Hears a Who", "Moby Dick", "Pink Honk-Honk"}

Are these sets any different?

...

Anyhoo...

What is $A - B$. Well we start with A= {"Moby Dick", "Horton Hears a Who", "Pink Honk-Honk"} and we must remove B which is {"Horton Hears a Who", "Moby Dick", "Pink Honk-Honk"}

So first we remove "Horton Hears a Who". That leaves us with {"Moby Dick", "Pink Honk-Honk"}

Then we remove "Moby Dick" that leaves us with {"Pink Honk-Honk"}.

Then we remove "Pink Honk-Honk" and that leaves us with {}.

So A - B = {}.

Now what is B - A? We start with B = {"Horton Hears a Who", "Moby Dick", "Pink Honk-Honk"} and we must remove A which is {"Moby Dick", "Horton Hears a Who", "Pink Honk-Honk"}.

First we remove "Moby Dick" which leaves us with {"Horton Hears a Who", "Pink Honk-Honk"}

Then we remove "Horton Hears a Who" which leaves us with {"Pink Honk-Honk"}.

Then we remove "Pink Honk-Honk" which leaves us with {}.

So B - A = {}.

What is {} - {}?

Well we start with nothing and we remove nothing leaving us with nothing.

(A-B) - (B-A) = {}-{} = {}.

By now I hope I hammered it home that A = B = {books in my living room} = {the same god-damned books no matter what f#@$ing order they are in}.

So if A = {1,2} and B = {1,2} then {1,2} = {2,1} because order doesn't f@#$ing matter, so A = B.

So (A-B) - (B-A) = (A-A) - (A-A) = $\emptyset -\emptyset = \emptyset$.

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Just elaborating on comments and existing answers, to maybe clarify your understanding of why order of elements doesn't matter in sets: Note that if $ A = \{1,2\}$ and $ B = \{2,1\}$ we have that $A \subset B$ since $\forall x \in A$ $x\in B$ and similarly $B \subset A$.

Thus $A=B$. (this is pretty much the axiom of extensionality in the ZFC axiom scheme)

Then of course that $A / B = \emptyset$

So $(A/B)/(B/A) = \emptyset$

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