Does there exist an operation or function which takes an operation and a set and returns the set with the operation applied to all of its elements?

I could just define the function $$F\left( n,\circ, \bigcup_i\left\{ x_i \right\}\right)=\bigcup_i\left\{n\circ x_i \right\}$$

but I imagine that there already exists some similar notation. For example, one can scale a vector $$n\langle a,b,c\rangle=\langle na,nb,nc\rangle$$ so why couldn’t one “scale” a set $$n\{a,b,c\}=\{na,nb,nc\}$$ in a similar manner?

I have seen some authors notate this kind of math by simply replacing the argument of a binary relation on two numbers with a set and a number (e.g., $\mathbb{Q}+1$).

  • $\begingroup$ If $A$ is the algebraic thing and $a \in A$ with $B \subseteq A$ you often see something like $a*B$ for $\{a*b: b \in B\}$. Replace the star by $+$, juxtaposition, $\circ$, etc. for your needs. $\endgroup$
    – Randall
    Sep 2, 2017 at 22:34
  • $\begingroup$ Beware of drawing bad conclusions like $|B| = |a*B|$ without actually proving them. In your example, what if $na=nb$? Then that set actually has (at most) two elements. $\endgroup$
    – Randall
    Sep 2, 2017 at 22:37

1 Answer 1


If f is a function and A a set, a common notation is the
set extenstion of f, f(A) = { f(x) : x in A } which is
also written as f[A].

  • $\begingroup$ So I would have to define a new function; there is no notation that already exists? $\endgroup$ Sep 3, 2017 at 0:18
  • $\begingroup$ I have shown existing notation. You are free to invent a functional, O(f) or O_f. $\endgroup$ Sep 3, 2017 at 6:05
  • $\begingroup$ So sorry—I meant without elsewhere defining $f$. $\endgroup$ Sep 3, 2017 at 6:08
  • $\begingroup$ Or is it okay to just say something like $2+\{1,2,3\}$? $\endgroup$ Sep 3, 2017 at 6:09
  • $\begingroup$ Yes you can as was pointed out earlier but only because + is a binary operator and because + is defined. No you cannot use undefined things. $\endgroup$ Sep 3, 2017 at 7:54

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