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The question is derived from this question I encountered:

Let $A$ be a set, and let there be a function $f: A \rightarrow A$, so that for every $a \in A$, $f(a) \neq a$.

Define $S=\{X \subseteq A: X \cap f(X)=\emptyset \}$. Show that if a chain $C=\{X_a: a \in I \}\subseteq S$, then $\cup C \in S$.

However, the meaning of $\cup C$ is unclear to me. $C$ is a set of sets, and $\cup C$ is a set made of the union of these sets? So it's a single set?

Thanks for any assistance!

Git Gud
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dsfsf
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  • Your intuitive interpretation is correct. Another way to look at it is to think of $\cup C$ as the set whose elements are exactly the elements of each $X_a$ grouped together.

    More formally and by definition one has, given a set (of sets if you wish) $C$ $$\cup C=\left{c\colon \exists Y\in \mathcal P(C)(c\in Y)\right}.$$

    In this particular example of $C$ the RHS of the above equality can be rewritten as ${c\colon \exists a\in I\left(c\in X_a\right)}$.

     – Git Gud Jun 21 '14 at 18:17

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Note that, by definition, $x \in \bigcup C$ iff there is some $y \in C$ such that $x \in y$. In your specific example, $\bigcup C$ consists precisely of those elements contained in some $X_a$, where $a \in I$.

So, if for example $C = \{X_1,X_2,X_3\}$, where $X_1 = \{1\}$, $X_2 = \{1,2\}$, $X_3 = \{3,4\}$, then $\bigcup C = \{1,2,3,4\}$.

Stefan Mesken
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