MHB Define Sets $\{x,y\}$ and $x \cup y$

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The discussion focuses on defining the sets $\langle x,y \rangle$ and their unions. It establishes that $\langle x,y \rangle = \{ \{x\}, \{x,y\} \}$, leading to $\bigcup \langle x,y \rangle = \{x,y\}$. The key point is that $\bigcup \bigcup \langle x,y \rangle = x \cup y$, which is justified by the definition of generalized union. This definition generalizes the union operation to potentially infinite sets. The clarification of these concepts enhances the understanding of set operations in mathematical contexts.
evinda
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Hey! (Wave)

If $x \neq y $, define the sets $\bigcup \langle x,y \rangle , \bigcup \bigcup \langle x,y \rangle$.

According to my notes, it is like that:

$$\langle x,y \rangle= \{ \{x\}, \{x,y\} \} $$

$$ \bigcup \langle x,y \rangle=\{x,y\} $$

$$ \bigcup \bigcup \langle x,y \rangle=x \cup y$$

Why is it $ \bigcup \bigcup \langle x,y \rangle=x \cup y$ and not $\bigcup \bigcup \langle x,y \rangle=\{x,y\}$ ? (Thinking)
 
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It follows from the definition of generalized union that $\bigcup\{A_1,\dots,A_n\}=A_1\cup\dots\cup A_n$. I would claim that this equality is the idea behind the definition of the generalized union, and this definition is a generalization to the case when the set is not necessarily finite.
 
Evgeny.Makarov said:
It follows from the definition of generalized union that $\bigcup\{A_1,\dots,A_n\}=A_1\cup\dots\cup A_n$. I would claim that this equality is the idea behind the definition of the generalized union, and this definition is a generalization to the case when the set is not necessarily finite.

I understand..Thank you very much! (Smile)
 
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