730: Union of Set

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definition of union of set

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note

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Starting Context

• The reader knows a definition of set.

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Target Context

• The reader will have a definition of union of set.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the sets }\}$
$\ast \cup S$: $\in \{\text{ the sets }\}$
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Conditions:
$p\in \cup S$
$⟺$
$\mathrm{\exists }{p}^{\prime }\in S(p\in p^{\prime })$
//

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2: Natural Language Description

For any set, $S$, the set, $\cup S$, such that $p\in \cup S$ if and only if $\mathrm{\exists }{p}^{\prime }\in S(p\in p^{\prime })$

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3: Note

$\cup S$ is indeed a set in the ZFC set theory by the union axiom.

Some frequently seen expressions like $S_1\cup S_2$ and ${\cup }_{\alpha \in A}{S}_{\alpha }$ are indeed $S_1\cup S_2:=\cup \{S_1,S_2\}$ and ${\cup }_{\alpha \in A}{S}_{\alpha }:=\cup \{S_{\alpha }|\alpha \in A\}$: when $A$ is a set, $\{S_{\alpha }|\alpha \in A\}$ is a set by the replacement axiom.

The special case of ${\cup }_{\alpha \in A}{S}_{\alpha }$ where $A=\{p\}$cannot be expressed as $\cup S_p$, because ${\cup }_{\alpha \in A}{S}_{\alpha }$ is $\cup \{S_p\}=S_p$ but not $\cup S_p$.

$\cup \mathrm{\varnothing }=\mathrm{\varnothing }$, because no element, $p$, cannot be in $\cup \mathrm{\varnothing }$, because there is no $p^{\prime }\in \mathrm{\varnothing }$ such that $p\in p^{\prime }$.

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References

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