Disjoint Union: Disjoint Unions
Definition
Let $\left\{ X_{\alpha} \right\} _{\alpha\in A}$ be an arbitrary index family. Then, the set of ordered pairs defined as follows is called the $\left\{ X_{\alpha}\right\}$ disjoint union.
$$ \bigsqcup \limits_{\alpha \in A} X_{\alpha} := \left\{ (x,\alpha)\ |\ x\in X_{\alpha},\ \alpha \in A \right\} $$
Explanation
Instead of $\bigsqcup$, $\amalg$, $\biguplus$, etc., are also used. Note that $\amalg$ is not the capital Pi $\Pi$. It is the mirrored version of it.
Even though they are different, to distinguish elements that look the same1, when forming the union, information about which set an element belongs to is added. For example, let’s say the set of students in class 1 is $X_{1}=\left\{ \right.$Kim Cheolsu, Kim Younghee, Park SuCheol, Lee HeeYoung$\left. \right\}$, and the set of students in class 2 is $X_{2}=\left\{ \right.$Kim Cheolsu, Kim Younghee, Kwon Hyunsu, Choi Changsik$\left. \right\}$. Then, Kim Cheolsu and Kim Younghee from class 1 and class 2 are clearly different people but appear to be the same. Thus, if we simply form the union, it might not accurately represent the actual union as $\left\{\right.$Kim Cheolsu, Kim Younghee, Park SuCheol, Lee HeeYoung, Kwon Hyunsu, Choi Changsik$ \left. \right\} \ne X_{1} \cup X_{2}$. However, if we form the disjoint union of $X_{1}$ and $X_{2}$,
$$ \bigsqcup \limits_{i=1,2} X_{i} =\left\{(\text{김철수},1), (\text{김영희},1), (\text{박수철},1), (\text{이희영},1), (\text{김철수},2), (\text{김영희},2), (\text{권현수},2), (\text{최창식},2) \right\} $$
it clearly indicates which class each belongs to, so two different elements aren’t treated as the same. If you understand the concept well, you should see that the following equation holds.
$$ \mathbb{R}^2 = \bigsqcup_{\alpha \in \mathbb{R}} \mathbb{R}_{\alpha} $$
Of course, this is meant to discuss abstract precision, so in practice, one wouldn’t mark it this complicatedly but think of it as forming a union without overlaps. That is, for each $\alpha \in A$, think of the following natural mapping as treating it as $x = (x,\alpha)$.
$$ \iota_{\alpha} : X_{\alpha} \hookrightarrow \bigsqcup \limits_{\alpha \in A}X_{\alpha} \quad \text{by } x \mapsto (x,\alpha) $$
Considering $X_\alpha$ and $\iota_\alpha (X_\alpha)$ the same. After all, the reason for considering the disjoint union is to say, “Kim Cheolsu from class 1 is a different person from Kim Cheolsu from class 2~ Don’t get confused,” so, in essence, it’s handled as follows.
$$ \bigsqcup \limits_{\alpha \in A} X_{\alpha} \approxeq \bigcup \limits_{\alpha \in A} X_{\alpha} $$
Or have the same name ↩︎