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I have encounter this example in the notes, but not sure what did it mean.

$\cup_{S \in C} S = \emptyset \cup \{ \emptyset \}=\{\emptyset\}$ where $C= \{\emptyset,\{\emptyset\}\}$

Is this means union set "S" itself and "S" is in the "C" set?

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  • $\begingroup$ The notation means "The union over all $S$ in $C$ of $S$". $\endgroup$ Commented Dec 15, 2012 at 11:12

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The notation $\bigcup_{S\in C}S$ means the union of all of the sets that are members of $C$. In this problem $C=\big\{\varnothing,\{\varnothing\}\big\}$, so as $S$ runs over the elements of $C$ it assumes just two values, $\varnothing$ and $\{\varnothing$. Thus,

$$\bigcup_{S\in C}S=\underbrace{\varnothing}_{\text{when }S=\varnothing}\cup\underbrace{\{\varnothing\}}_{\text{when }S=\{\varnothing\}}=\{\varnothing\}\;,$$

where the last step is because $\varnothing\cup A=A$ for any set $A$.

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The definition of $\cup_{S \in C} S$ is: $\{x|\exists S\in C[x\in S]\}$

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    $\begingroup$ Just to confuse the OP: This is often simply written as $\bigcup C$. $\endgroup$ Commented Dec 15, 2012 at 11:19
  • $\begingroup$ yes. I saw this notation in my formal logic course. $\endgroup$
    – Amr
    Commented Dec 15, 2012 at 11:21
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In general $\bigcup_{S \in C} S$ denotes the union of all sets belonging to the collection $C$, i.e., the collection of all objects that belong to at least one set in $C$.

If $C = \{ \emptyset , \{ \emptyset \} \}$, then $\bigcup_{S \in C} S = \emptyset \cup \{ \emptyset \} = \{ \emptyset \}$.

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