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There are 5 sets A,B,C,D and E.

A,B,C,D are all different from eachother.

E contains elements of at most two sets of A,B,C,D.

How can I write that with math symbols without listing all possible cases?

Normal language is allowed but should be minimized.

Example what E could look like:

$E=\{a_m,...,a_n,b_j,...,b_k\}$

Or $E=\{\}$

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What I think the condition you're describing means is

$$ \Bigl| \{X\in\{A,B,C,D\}\mid X\cap E\ne \varnothing\}\Bigr| \le 2 $$

But this would require your example $E$ to be $\{a_m,...,a_n,b_j,...,b_k\}$, rather than $\{\{a_m,...,a_n\}\{b_j,...,b_k\}\}$.

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  • $\begingroup$ can you explain why $X \cap E\neq\emptyset$? Because E can be an empty set, and any intersection with $\emptyset$ will also be $\emptyset$ or did i get something wrong? $\endgroup$ – SAJW Oct 23 '16 at 4:46
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    $\begingroup$ @saturatedexpo: That's fine, because if $E$ is the empty set then the set-builder condition fails for all possible $X\in\{A,B,C,D\}$ in which case the set $\{X\in\{A,B,C,D\}\mid X\cap E\ne \varnothing\}$ is empty and $\Bigl| \{X\in\{A,B,C,D\}\mid X\cap E\ne \varnothing\}\Bigr| = 0$, which is indeed less than or equal to $2$. $\endgroup$ – Richard Ambler Oct 23 '16 at 5:17
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$(E \subseteq \{A, B, C, D\}) \land (|E| \le 2) $

Translation:

E is a subset of {A, B, C, D}, so that every element of E is one of A, B, C, D; and ($\land$) the number of elements of E (written "|E|") is at most two.

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  • $\begingroup$ hi, does this account for the case that "some a in A are in E, but not all" AND "some b in B, but not all" for example? Because that was the point where i thought cardinality would go senseless. $\endgroup$ – SAJW Oct 23 '16 at 4:18
  • $\begingroup$ I would understand the (not very well phrased) question to mean $E\subseteq A\cup B\cup C\cup D$, not $E\subseteq\{A,B,C,D\}$. $\endgroup$ – hmakholm left over Monica Oct 23 '16 at 4:21
  • $\begingroup$ @HenningMakholm yes that's right. but then the sweet cardinality <3 can't be right or? (sweet as in super crisp) Should i edit the Question? $\endgroup$ – SAJW Oct 23 '16 at 4:22
  • $\begingroup$ $\{A, B, C, D\}$ is a set of sets, not a set of the elements of $A$, $B$, $C$ and $D$... Marty Cohen's interpretation is correct but the question said that $E$ should contain the elements of at most two of the sets $A$, $B$, $C$ and $D$. $\endgroup$ – Richard Ambler Oct 23 '16 at 4:29
  • $\begingroup$ [Note: Following some objections, I post the following as comment which I had earlier posted as answer] Your words E contains elements of at most two sets of A,B,C,D . are good enough. Symbols should bring clarity. If they don't don't use them. One can write a convoluted symbolic expression that expresses this condition and nobody will be any wiser. Try to write the following symbolically: The factorization of number n involves at most two primes between 13 and 47. $\endgroup$ – P Vanchinathan Oct 23 '16 at 14:28

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