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Editing since this seems to be unclear: I ONLY have the data listed below. I do not have any information on the underlying elements. For example, I know A&B = 14, but I don't know which 14 elements match, or the values of those 14 elements.

I am trying to determine if the following is possible.

You have 3 sets A, B, C

You have the Size of A, Size of B, and Size of C.

You also have the intersect of A&B, intersect of A&C, and the intersect of B&C.

Is it possible to determine the intersect A&B&C?

Example:

size A = 25

size B = 30

size C = 22

A&B = 14

A&C = 10

B&C = 9

What is A&B&C?

Also, if this is possible, can it also be done on greater number of sets. For example, A,B,C,D given you have all intersect pairs?

Side note: I have been able to prove the following, but haven't been able to apply it to >2 sets.

Size A = 10

Size B = 20

AUB = 25

A&B = (size A + size B) - AUB

2 Answers 2

1

No. Consider the following examples:

$A = \{1,2\}$, $B = \{2,3\}$, $C = \{1,3\}$.

$A = \{1,2\}$, $B = \{1,3\}$, $C = \{1,4\}$.

In each, $|A| = |B| = |C| = 2$ and $|A\cap B| = |B\cap C| = |A\cap C| = 1$. But in the first example, $|A\cap B\cap C| = 0$, while in the second example, $|A\cap B\cap C| = 1$.

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Yes. $$A \cap B \cap C = (A \cap B) \cap C = (A \cap C) \cap (B \cap C)$$

So you just need to find the intersection of $A \cap C$ and $B \cap C$. Note that the sizes are not necessary info here.

In words:

You know which elements $A$ and $B$ have in common. You know which elements $B$ and $C$ have in common. You want to know which elements $A$ and $B$ and $C$ all have in common. So you just need to take the common elements of your intersections $A \cap B$ and $B \cap C$. The first gives you everything that's in both $A$ and $B$. The second gives you everything that's in both $B$ and $C$. So if you intersect them, you get everything that's in both $A$ and $B$ that's also in both $B$ and $C$, i.e., everything in $A$ and $B$ and $B$ and $C$, i.e., everything in $A$ and $B$ and $C$.

It's also possible to show $(A \cap C) \cap (B \cap C) = A \cap B \cap C$ in the "standard" way of showing each side is a subset of the other:

Let $x \in (A \cap C) \cap (B \cap C)$. Then $x \in A \cap C$ and $x \in B \cap C$. Since $x \in A \cap C$ then $x \in A$ and $x \in C$. Since $x \in B \cap C$ then $x \in B$ and $x \in C$. So we have $x \in A$, $x \in B$, and $x \in C$. Therefore $x \in A \cap B \cap C$. This shows that $(A \cap C) \cap (B \cap C) \subseteq A \cap B \cap C$. The other direction is left as an exercise for the reader.

Yes, this can be extended. If your collection of sets is finite, and if you know all of the paired intersections, then most generally you will have \begin{align*}\bigcap_{k=1}^n A_k &= A_1 \cap A_2 \cap A_3 \cap \cdots \cap A_{n-1} \cap A_n \\ &= (A_1 \cap A_2) \cap (A_2 \cap A_3) \cap (A_3 \cap A_4) \cap \cdots \cap (A_{n-1} \cap A_n) \end{align*}

Note that if you have an even number of sets in your collection then you don't need the "overlap" (but it's still not wrong to have it). $$ \bigcap_{k=1}^{2m} A_k = (A_1 \cap A_2) \cap (A_3 \cap A_4) \cap (A_5 \cap A_6) \cap \cdots \cap (A_{2m-1} \cap A_{2m})$$

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    I have A∩C and B∩C. But how would I calculate (A∩C) ∩ (B∩C) without having the underlying data? The problem isn't how to calculate an intersection of 3 sets, it's how to do it only with the data listed above. – 2017-02-14
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    I know the number of elements they have in common. I do not know which elements they have in common. – 2017-02-14
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    Would the downvoter care to explain? – 2017-02-14
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    @user2263572, what do you mean you have $A \cap C$ and $B \cap C$ but you don't have the "underlying data"? "I have $A \cap C$ and $B \cap C$" means "I know what elements are in $A \cap C$ and $B \cap C$." Please rephrase in your post if you don't actually have those elements. – 2017-02-14
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    Downvote wasn't me, but I edited my question to explain in better detail. From an outside source I am given A∩C and B∩C. I do not calculate these on my own. – 2017-02-14
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    I downvoted. The question is not clearly worded, but I think it's clear from context that the OP wants to find the cardinality of the triple intersection, given only the *cardinalities* of the single and double intersections. This is not possible, as I show in my answer. – 2017-02-14
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    @AlexKruckman, I disagree that it's "clear" since I chose to interpret "You have the intersect of **A&B**..." at face value and took the rest of it as an abuse of notation, something which I've commonly seen on this site. – 2017-02-14
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    I agree that the problem statement is incorrect (it should say "You also have the *sizes* of the intersect of A&B, intersect of A&C, and the intersect of B&C"). But then you look at the example instance of the problem, which contains lines like "A&B = 14". This is obviously meant to be a statement about the cardinality of the intersection. This is the context I'm referring to. – 2017-02-14
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    See also the line "A&B = (size A + size B) - AUB" – 2017-02-14
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    @user2263572 "From an outside source I am given $A \cap C$ and $B \cap C$" still means you know which *elements* are in $A \cap C$ and $B \cap C$. It doesn't matter whether this knowledge comes from an outside source or not. If you know $A \cap C$ and $B \cap C$ then you automatically know $(A \cap C) \cap (B \cap C)$ simply by looking at the elements that $A \cap C$ and $B \cap C$ have in common. – 2017-02-14
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    @AlexKruckman, I know which context you're referring to; that's exactly what I was talking about when I mentioned an abuse of notation. – 2017-02-14