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Hi, I have what I hope is a very simple question related to unfamiliar notation. I am looking through a maths paper on a topic related to set theory which contains a symbol,


and I would like to know how, if at all, it differs from the typical


symbol in terms of its meaning.
The context leads me to believe that it does not in fact differ at all but since I don't even know the name of the symbol other than the latex id that I looked up, I can't seem to confirm that suspicion. Cheers

edit: it seems that the latex renderer also does not know about this obscure symbol '\cupplus' but '\uplus' does work.

edit2: thanks for all the replies! however, since everyone commented rather than providing an answer it seems I cannot grant the coveted 'answer' status to anyone. the disjoint union makes the most sense.

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It is sometimes used for disjoint union, particularly by combinatorists. – Dan Petersen Apr 16 '10 at 8:19
And also for the similar notion of multiset sum, see – user2734 Apr 16 '10 at 8:25
I don't know much about set theory, but I can definitely say I have found that symbol somewhere, meaning "disjoint union". The disjoint union of (not necessarily disjoint sets) $A$ and $B$ is by definition the usual union between $A\times \{ 0\}$ and $B \times \{1\}$. – Qfwfq Apr 16 '10 at 9:16
I too have seen this symbol being used to denote disjoint unions. However, I believe it is not quite the same as the coproduct, where one constructs disjoint union out of arbitrary sets. This one here only wants to make the fact apparent that both sets are disjoint as subsets of some set in the usual sense. – efq Apr 16 '10 at 10:02
It's sometimes used to mean "$A \cup B$ (and $A \cap B = \varnothing$)" just like ${\oplus}$ in linear algebra means "$A + B$ (and $A \cap B = \{0\}$); both are horrible abuses of notation -… – François G. Dorais Apr 16 '10 at 13:33
up vote 3 down vote accepted

For the sake of giving a response to tag as an answer:

Some authors use this to denote a disjoint union of sets, i.e., "$A \cup B$ and $A \cap B = \emptyset$."

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