Prove that $\bigcup \{A,B \} = A \cup B$. [closed]

Prove that $$\;\bigcup\big\{A,B\big\} = A \cup B$$.

I am trying to work on this problem here. But I just do not know where to start. So I know that the union of set {A,B} is {A,B}. And the union of Set A and B we also get {A,B} for any common elements. How would I show this proof wise

• Start with the definitions. Once you have those, a common way to prove two sets are equal is to prove that each is a subset of the other. Commented Feb 20, 2023 at 19:40
• Please show your attempt or thoughts. Commented Feb 20, 2023 at 19:40
• Am I the only one who has no idea what this notation means? Commented Feb 20, 2023 at 20:10
• @AndrewZhang For a set $X$ we have $\bigcup X = \{x \mid \exists A \in X : x\in A\}$. Commented Feb 20, 2023 at 20:26
• @AndrewZhang It's the standard notation for the union of a set of sets after one adopts the Axiom of Union Commented Feb 20, 2023 at 20:46

Let $$x \in \bigcup \{ A, B \}$$ be arbitrary. This means there exists $$C \in \{ A, B \}$$ such that $$x \in C$$. We can simply enumerate to find that either $$x \in A$$ or $$x \in B$$, and this implies $$x \in A \cup B$$. So $$\bigcup \{ A, B \} \subseteq A \cup B$$.

Let $$x \in A \cup B$$ be arbitrary. This means either $$x \in A$$ or $$x \in B$$. If $$x \in A$$, then because $$A \in \{ A, B \}$$, it must be that $$x \in \bigcup \{ A, B \}$$. If $$x \in B$$, then because $$B \in \{ A, B \}$$, still $$x \in \bigcup \{ A, B \}$$. So $$A \cup B \subseteq \bigcup \{ A, B \}$$.

Therefore, $$\bigcup \{ A, B \} = A \cup B$$.

So I know that the union of set {A,B} is {A,B}.

It seems like you are not clear on what $$\bigcup \mathcal S$$ means when $$\mathcal S$$ is a set of subsets of $$X$$. The definition is:

$$\bigcup \mathcal S=\{x\in X\mid \exists S\in\mathcal S \text{ such that } x\in S\}$$

So while $$\bigcup\{\{A,B\}\}=\{A,B\}$$, it is not true that "$$\bigcup\{A,B\}=\{A,B\}$$."

I recall the helpful thing someone said to me the first time I was learning it: they said "it's like the union symbol breaks down the walls between sets in $$\mathcal S$$."

So, for example, $$\bigcup \{\{a,b\},\{c,d\}\}$$ dissolves the inner braces and you just get $$\{a,b,c,d\}$$.

As far as I know, $$A\cup B$$ is just a convenient shorthand for $$\bigcup \{A,B\}$$ in order to think of it as a binary operator. I'm not aware of any definition that defines the two differently from each other.