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I don't have a lot of experience with set theory, as I suspect this question will make clear!

As the title says, I'm interested in expressing the non-intersecting sections of three sets using a single set operation.

What I mean is, how would you express the gray shaded regions in the following Venn Diagram?

Venn Diagram with non-intersecting areas shaded

My best guess is that this has to do with complements... something like

((A ∪ B) ∪ (A ∪ C) ∪ (B ∪ C))C

...maybe? But I am concerned that this complement would also include the universe, U, which I don't want to be part of the selection. I also suspect that my notation is wrong, so apologies if that's the case.

Thanks for any help!

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    $\begingroup$ Your expression is $(A \cup B \cup C)^c$, exactly the stuff outside the whole figure. $\endgroup$ – Ross Millikan Nov 12 '13 at 18:56
  • $\begingroup$ @RossMillikan Oh, right, thanks for pointing that out! I think what I meant to say is ((A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C))c. In other words, trying to express the complement of the white-shaded region. Does that make more sense (even if it's still wrong)? $\endgroup$ – user108338 Nov 12 '13 at 19:58
  • $\begingroup$ That works fine except it includes everything outside the diagram as well. You can intersect it with $A \cup B \cup C$ to solve that. $\endgroup$ – Ross Millikan Nov 12 '13 at 20:01
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So you want the part $A$ without those parts in $B$ and $C$; hence you want $$A \backslash (B \cup C)$$

Likewise you want $B$ without either $A$ or $C$, hence you also-want $$B \backslash(A \cup C)$$; and similarly you want $C$ without $A$ or $B$; hence also desire $$C \backslash(A \cup B)$$

Now overall you want any of the above, hence, you union to obtain $$\boxed{A \backslash (B \cup C) \;\;\cup\;\; B \backslash(A \cup C) \;\;\cup\;\; C \backslash(A \cup B)}$$

Hope that helps!


Note:: $X \backslash Y$ means the set $X$ without the elements in the set $Y$; read "X less Y".

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  • $\begingroup$ This seems like the most straightforward method, though I think @kbball's method makes a lot of sense, too. $\endgroup$ – user108338 Nov 12 '13 at 21:23
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$(A-B-C)\cup (B-A-C) \cup (C-A-B)$

Or you could do:

$(A \cup B \cup C) -(A \cap B)-(A \cap C)-(B \cap C)$

In either case, you want to use subtraction of sets. In case you are not familiar, here is an example:

Consider the set $D=\{1,2,3,4,5\}$ and $E=\{3,4,5,6,7\}$.

Then the set $(D-E)=\{1,2\}$ and the set $(E-D)=\{6,7\}$.

Hence, $(D-E)\cup (E-D)=\{1,2\} \cup \{6,7\} = \{1,2,6,7\}$.

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  • $\begingroup$ Thanks! This makes a lot of sense! $\endgroup$ – user108338 Nov 12 '13 at 21:25
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Hint: The shaded $A$ region is $A \cap (B \cup C)^c$.

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I think ((A ∪ B) ∪ (A ∪ C) ∪ (B ∪ C))$^c$ would just give $(A \cup B \cup C)^c$

how about

$ (A \cap (B \cup C)^c) \cup (B \cap (A \cup C)^c) + (C \cap(A \cup B)^c)$

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  • $\begingroup$ Wouldn't A ∪ (B ∪ C)c just be A, since we start with all of A and then "add" to it the gray section of A, leaving us with all of A? $\endgroup$ – user108338 Nov 12 '13 at 20:04

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