How would one denote the set of all subsets of $A$ which have size $2$?

Would $$\binom{A}{2}$$

be a good idea?

  • $\begingroup$ I believe the notation is $[A]^2$. Or atleast, we are using it in our set theory class. $\endgroup$ – RKD Feb 24 '12 at 17:21
  • $\begingroup$ With nonstandard notation (at least I believe this is nonstandard notation), I think that being as explicit as possible as helpful. Thus, I would write $$\{X \subset A : |X| = 2\}.$$ Your notation, however, does emphasize the size of such a set which is a great quality in a notation. $\endgroup$ – JavaMan Feb 24 '12 at 17:23
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    $\begingroup$ Dear stefan: It wouldn't be a good idea: it would be a terrific idea! $\endgroup$ – Pierre-Yves Gaillard Feb 24 '12 at 17:23
  • $\begingroup$ @JavaMan I am not sure whether your notation is First Order Notation I suspect it is Second Order which may be not feasible in some cases. Asaf Karagila had given nice first order definition along with short hand notation. I think $\{x \in 2^{|2|}:|x| =2\}$ might be first order notation but I am not completely sure. $\endgroup$ – Trismegistos Feb 24 '12 at 18:02
  • $\begingroup$ I made a mistake in my notation I meant $\{X\in A^{|2|}:|X|=2\}$ $\endgroup$ – Trismegistos Feb 24 '12 at 18:15

$\dbinom A 2$ is standard notation for the set of all size-$2$ subsets of a set $A$, in the usage of combinatorialists.

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    $\begingroup$ It is a great notation. However, heavily used in inline math it tends to either get cramped or to push the lines apart. So while the mathematician in me approves, my inner typographer does not. $\endgroup$ – Harald Hanche-Olsen Feb 24 '12 at 17:38
  • $\begingroup$ @Harald: There is still some effect of you use a smaller version such as $\tbinom A 2$ but I find it quite reasonable. $\endgroup$ – Henry Feb 24 '12 at 18:06
  • $\begingroup$ @Henry: I disagree. The letter A is too small. Maybe that is because I am getting a bit old? In the same vein, I dislike text-mode fractions except for simple numerical fractions like $\frac56$. $\endgroup$ – Harald Hanche-Olsen Feb 24 '12 at 18:28
  • $\begingroup$ I always thought this was the number of unordered pairs from $A$. Which never made real sense in case $A$ was not finite. $\endgroup$ – Asaf Karagila Feb 24 '12 at 19:26
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    $\begingroup$ The number of unordered pairs from $A$ is $\dbinom{|A|}{2}$ (where $|A|$ is the number of members of the set $A$). The set of unordered pairs from $A$ is $\dbinom{A}{2}$. $\endgroup$ – Michael Hardy Feb 25 '12 at 4:43

In set theory it can be often denoted as $[A]^2=\{\{a,b\}:a,b\in A, a\neq b\}$.

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    $\begingroup$ Add condition that $a\neq b$. $\endgroup$ – Thomas Andrews Feb 24 '12 at 17:49

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