731: For Set, Union of Power Set of Set Is Set

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description/proof of that for set, union of power set of set is 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: Proof
• 4: Note

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Starting Context

• The reader knows a definition of power set of set.
• The reader knows a definition of union of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set, the union of the power set of the set is the 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:
$S$: $\in \{\text{ the sets }\}$
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Statements:
$\cup Pow(S)=S$
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2: Natural Language Description

For any set, $S$, $\cup Pow(S)=S$.

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3: Proof

Whole Strategy: Step 1: see that each element of $\cup Pow(S)$ is contained in $S$; Step 2: see that each element of $S$ is contained in $\cup Pow(S)$; Step 3: conclude the proposition.

Step 1:

Let $p\in \cup Pow(S)$ be any.

$p$'s being in the union of $Pow(S)$ means that $p$ is in an element of $Pow(S)$, which (the element) is a subset of $S$, so, $p$ is in a subset of $S$, which implies that $p\in S$.

So, $\cup Pow(S)\subseteq S$.

Step 2:

Let $p\in S$ be any.

$\{p\}\in Pow(S)$, so, $p$ is in an element of $Pow(S)$, so, $p\in \cup Pow(S)$.

So, $S\subseteq \cup Pow(S)$.

Step 3:

So, $\cup Pow(S)=S$.

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4: Note

On the other hand, $Pow(\cup S)\ne S$ in general: while always $\mathrm{\varnothing }\in Pow(\cup S)$, $\mathrm{\varnothing }$ is not in $S$ in general.

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References

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