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I've tried to Google this by using a variety of search terms but I am not even sure exactly what to ask.

Q. What is the meaning of "$AB$" where $A$ and $B$ are sets?

I have looked up all the set operations listed on various websites explaining set theory operations and I can't find the meaning of simply writing the two set names concatenated together. I first encountered the "$AB$" expression in a textbook I am reading now and no explanation is given for its meaning.

Any explanation would be appreciated.

Context: a proof is given for the following statement.

For any event $A$ and $B$, $$ P(A \cup B) = P(A) + P(B) - P(AB) $$

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    $\begingroup$ Use in context? $\endgroup$
    – Eric Towers
    Oct 23, 2017 at 18:00
  • $\begingroup$ Please provide more context. $\endgroup$
    – José Carlos Santos
    Oct 23, 2017 at 18:00
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    $\begingroup$ The expression can take on different meanings, just like some authors use $A + B$ to denote $A \cup B$. Could you provide us some context, i.e where you saw the expression? $\endgroup$
    – rubikscube09
    Oct 23, 2017 at 18:00
  • $\begingroup$ It needs context. It may mean the set containing all products $ab$ with $a \in A, b \in B$ (whatever product means). It could mean the concatenation of strings, etc. In general, $A \square B = \{ a \square b | a \in A, b \in B \}$. Presumably there is some implied $\square$ operation in context. $\endgroup$
    – copper.hat
    Oct 23, 2017 at 18:01
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    $\begingroup$ @copper.hat And even then there are exceptions to that general interpretation, such as the product of ideals. Another reason why context is so important :). $\endgroup$
    – Erick Wong
    Oct 23, 2017 at 18:12

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It means "intersection." This goes back to old algebraic ideas in logic, that connect "$+$" with "$\cup$" and "$\times$" with "$\cap$."

. . . at least, that's what it means in this context. There are other contexts where it means something different - the other meanings I've seen are $\{ab: a\in A, b\in B\}$ when $A, B$ are sets of numbers (combinatorics), $\{a^\smallfrown b: a\in A, b\in B\}$ (also combinatorics), and union (model theory).

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  • $\begingroup$ Thank you. I went line by line through the proof with the understanding that AB is the same as A n B and everything made sense. $\endgroup$
    – BJDE
    Oct 23, 2017 at 18:15

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