For a set $S$, for "all possible subsets of $S$" you have $\mathcal{P}(S)$.

For a set $S$ consisting of sets, for "the union of all sets $T\in S$" you have $\bigcup_{T\in S}T$.

Is there a notation for "all possible unions of one or more sets $T\in S$"?

e.g. for $S=\{\{1,2\},\{3,4\},\{5\}\}$, such an "all possible unions" set would be $\{\{1,2\},\{3,4\},\{5\},\{1,2,3,4\},\{1,2,5\},\{3,4,5\},\{1,2,3,4,5\}\}$.


I don't think You can find a specific notation for it, but you can write it as set: $$ \big\{ \bigcup \limits_{T \in U} T : \; U \in \mathcal{P} (S) \big\} $$

  • $\begingroup$ I had not thought of that notation. Silly me! $\endgroup$ – JustAskin Mar 8 '14 at 11:30
  • $\begingroup$ There is no silliness in asking :) $\endgroup$ – wroobell Mar 8 '14 at 11:34
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    $\begingroup$ In order to get "all possible unions of one or more sets $T \in S$" you are going to have to additionally disallow $U = \varnothing$. $\endgroup$ – user642796 Mar 8 '14 at 15:56
  • $\begingroup$ @ArthurFischer For the set I defined in my original post, $\varnothing\not\in S$. $\varnothing\subset S$ but not element of. Wouldn't this mean that $\varnothing\not\in \big\{ \bigcup \limits_{T \in U} T : \; U \in \mathcal{P} (S) \big\}$ ? edit: Nevermind, $\varnothing\not\in S$ but $\varnothing\in\mathcal{P}(S)$ so you're right. $\endgroup$ – JustAskin Mar 9 '14 at 3:16
  • $\begingroup$ Of course, these are the same. Maybe even the second one is neater. As @Arthur Fischer pointed out it should be: $$ \big\{ \bigcup \limits_{T \in U} T : \; U \subset S \wedge U \neq \emptyset \big\}$$ $\endgroup$ – wroobell Mar 9 '14 at 10:09

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