Set-builder notation for unique values in a set

Assume a set which contains non-unique numbers. That is $$X = \{A \mid (\exists i,j)[A_i=A_j] \}$$. There is another set which contains the unique values from X with a weight associated with each unique value. For example, if $$X=\{10,10,20\}$$, then $$U=\{(10,0.5),(20,0.5)\}$$. I would like to use set-builder notation to write $$U$$. My initial thought is

$$U = \left\{ (A^\prime,W) \mid (\forall i \neq j)[A^\prime_i \neq A^\prime_j] \right\}$$ where $$|U| \leq |X|$$ (size of $$U$$ is definitely not larger than $$X$$) and $$W_i=\Sigma(???)/\Sigma(A_i)$$. I don't know how to write the nominator for $$W_i$$ to say "summation of those $$A_i$$s which are equal".

Any idea on how to properly express that? If you think $$X$$ and $$U$$ need corrections, fell free to propose that.

P.S: There is no "set-builder" tag and I can not add that.

• A set has no "duplicate" elements: thus, $X = \{ 10,10,20 \}$ is simply $X = \{ 10,20 \}$. If we want "duplicate" elements with "weights" maybe a sequence $(A_i)$ with weights $(w_i)$ and thus pairs $(A_i, w_i)$. May 3, 2022 at 15:52
• OK. Then $U$ is a set of pairs. Right? May 3, 2022 at 21:46

Here is one approach, using disjoint unions. Let's say you have some sets $$X_1,X_2,X_3,\dots$$ of numbers (say integers). While each set $$X_i$$ doesn't contain any repeated elements (by definition), we may certainly have $$X_i \cap X_j \neq \emptyset$$ for distinct $$i$$ and $$j$$. Then, consider the set: $$X = \bigsqcup_i X_i.$$ The elements of $$X$$ look like $$(x_i,i)$$, where $$x_i \in X_i$$. Why is this useful? Well, for any $$x \in \bigcup_i X_i$$, we can determine its weight $$w(x)$$ by: $$w(x) := \#\{ i : (x,i) \in X\}$$ Since you asked for the weights in set-builder notation, we can write it as: $$W := \Big\{(x,w(x)) : x \in \bigcup_i X\Big\}.$$ Here is an example to be concrete. Let $$X_1 = \{10,20\}$$, $$X_2 = \{10,20,25\}$$, $$X_3 = \{10,5\}$$. Then, we have: $$X = X_1 \sqcup X_2 \sqcup X_3 = \big\{(10,1),(20,1),(10,2),(20,2),(25,2),(10,3),(5,3)\big\}.$$ From here, we easily compute $$w(10) = 3$$, $$w(20) = 2$$, $$w(25) = 1$$, and $$w(5) = 1$$. In other words, $$W = \{(10,3),(20,2),(25,1),(5,1)\}$$
Hope this helps. (And if you like the weights as percentages instead of numbers, you can of course divide $$w(x)$$ by $$\sum_x w(x)$$, where of course the sum is over $$x \in \bigcup_i X_i$$)