I would like to denote the following set:

$$ \bigl\{\{a,b\}:a\in A\text{ and }b\in B\bigr\} $$

I have found this (Notation for unordered product of sets) but I do not know how this is answer my question. The answer says to use the notation $\binom{V}{2}$ but I have two sets $A$ and $B$ not only one $V$.


  • $\begingroup$ I don't know of a symbol for this. It's probably fine if you make up your own as long as the definition is clear. $\endgroup$ – Jair Taylor Feb 15 '16 at 3:20
  • $\begingroup$ why not use the description you gave: $\{\{a,b\}: a\in A, b\in B\}$? $\endgroup$ – Jorge Fernández Hidalgo Feb 15 '16 at 3:20
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    $\begingroup$ To begin with, what if $A\cap B\neq \emptyset$? If $x\in A\cap B$, you wouldn't write $\{x,x\}$ as a set, as that would simply be the set $\{x\}$. $\endgroup$ – JMoravitz Feb 15 '16 at 3:20
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    $\begingroup$ @JMoravitz I don't see any harm in this. $\endgroup$ – Stefan Mesken Feb 15 '16 at 3:22
  • $\begingroup$ Thank you all, I will just use this one $\{\{a,b\},a\in A\text{ and } b\in B\}$ then. $\endgroup$ – Chiba Feb 15 '16 at 3:34

Assuming that $A$ is disjoint from $B$, you could denote this as:

$$\left\{\{a,b\}~:~a\in A,b\in B\right\}=\binom{A\cup B}{2}\setminus \left(\binom{A}{2}\cup \binom{B}{2}\right)$$

I.e. the family of all subsets of size two of $A\cup B$ such that the elements are neither both from $A$ nor are both from $B$, using the notation that $\binom{X}{2} = \{E~:~E\subseteq X,~|E|=2\}$, the family of all subsets of size two of $X$.

Beyond that, this is not commonly used and I know of no symbol to denote this. Creating your own symbol to use so long as you define it in any paper you use it in for the reader is fine, or you can simply skip using a symbol altogether and just write it as the set in the way you already did in your question.

  • $\begingroup$ If you wish to fix the above for use with sets that are not necessarily disjoint, you could try playing with unioning the above with things like $\binom{A}{2}\setminus \binom{A\setminus B}{2}$ to add back in the pairs which were accidentally removed. $\endgroup$ – JMoravitz Feb 15 '16 at 3:35

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