📂Set Theory

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.

$$\bigsqcup \limits_{\alpha \in A} X_{\alpha} \approxeq \bigcup \limits_{\alpha \in A} X_{\alpha}$$