Addition of Sets which isn't union

Question

today a student asked me to prove $${A} \cup B \cup C = A+ B+ C- A\cap B - A\cap C$$

I really had no idea what precisely the "+" sign meant, they insisted, "You know you just add the sets together"; of course, they also had no textbook.

I assumed then that perhaps $$A\cup B = A+B - A\cap B$$ But I'm not really sure what "-" means here, normally I'd interchange that with set subtraction \ and I occasionally use + to mean $\cup$. In any case, I think + sort of gets extra copies of the parts they share. But I really have no idea, any clues.

Maybe the student meant $\left|A\cup B\cup C\right|=|A|+|B|+|C|-\left|A\cap B\right|-\left|A\cap C\right|$. That wouldn't be true unless $B\cap C\subseteq A$. — Thomas Andrews, Oct 31 '13 at 03:22
It' certainly unlikely to be true without some conditions, since the left side is symmetric in all three variables, but the right side you can only swap $B$ and $C$. — Thomas Andrews, Oct 31 '13 at 03:24
If there is a function from the space of the sets into the reals (v.g. cardinality, or probability, or lenght) that behaves like a measure... — Carlos Eugenio Thompson Pinzón, Oct 31 '13 at 03:26

score 4 · Accepted Answer · answered Oct 31 '13 at 03:21

It looks like the student was talking about cardinalities of sets rather than the sets themselves. For instance, it holds true that $$|A \cup B| = |A| + |B| - |A \cap B|$$ But in this scenario the first equation still doesn't hold, because $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$ But if a student is prepared to say "you know, just add the sets" then it's not entirely infeasible that they just omitted/forgot/incorrectly copied the last two terms.

(There's no obvious definition of $+$ that would make your first equation hold, where by 'obvious definition' I mean either with cardinalities as above, or where $+$ means disjoint union.)

score 1 · Answer 2 · edited Apr 20 '14 at 01:04

Before looking it up just now, I didn't realize you couldn't have duplicate elements in a set. Intuitively "$A\cup B\cup C = A \cup B - A\cap B - A\cap C$" made a lot of sense to me because of this. If you were under the misconception that sets could have multiple copies of the same element (as I was just a couple of minutes ago), then it's not much of a stretch to also see the union of A and B ($A\cup B$) as just grouping all of the elements into one set (A+B) and filtering out the duplicates ($-A\cap B$).

I assumed that A + B is (informally) defined as simply conjoining two sets without regard to duplicity of elements.

First off: The operation "+" maps two sets or "setts" to a "sett", where a "sett" is essentially a set with duplicity allowed. So $\{1,2,3\} + \{2,3,4\} = \{\{1,2,2,3,3,4\}\}$.

And subtraction is simply your regular set subtraction expanded to operate on "setts".

From this definition you can see that $A\cup B + A\cap B = A+B$. So yes, essentially $A\cup B = A+B - A\cap B$.

You were exactly right with your guess that $A+B$ "sort of gets extra copies of the parts they share".

I could work on proving the student's original statement, but I'm sure people smarter than me can figure it out. Not to mention I didn't give very clear formal definitions for '+'.

Also, I apologize for my terrible formatting.

Sweet, thanks for the reference. Multisets are exactly what I was thinking of. — Luke, Apr 20 '14 at 01:59
@user144339 So if I understand you correctly, basically your answer is, "This student was probably thinking about multisets (or bags), instead of sets." See http://math.stackexchange.com/q/152223/11994 for a nice discussion of this topic. — MarnixKlooster ReinstateMonica, Apr 20 '14 at 05:50

Addition of Sets which isn't union

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