637: Quotient Set

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definition of quotient 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 equivalence relation on set.

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Target Context

• The reader will have a definition of quotient 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:
$S$: $\in \{\text{ the sets }\}$
$\sim$: $\in \{\text{ the equivalence relations on }S\}$
$\ast S/\sim$: $=\{p\in Pow(S)\setminus \mathrm{\varnothing }|\mathrm{\forall }{p}^{\prime },p^{″}\in p(p^{\prime }\sim p^{″})\wedge \mathrm{\forall }{p}^{\prime }\in p,\mathrm{\forall }{p}^{″}\in S\setminus p(\mathrm{\neg }{p}^{\prime }\sim p^{″})\}$
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Conditions:
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$S/\sim$ is well-defined, because it is defined based on the subset axiom with a legitimate formula. Whether $S/\sim$ partitions $S$ is another issue, but $S/\sim$ indeed partitions $S$ as is shown in Note.

In other words, the quotient set is the set of the equivalence classes of $S$ by $\sim$.

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2: Natural Language Description

For any set, $S$, and any equivalence relation on $S$, $\sim$, $S/\sim =\{p\in Pow(S)\setminus \mathrm{\varnothing }|\mathrm{\forall }{p}^{\prime },p^{″}\in p(p^{\prime }\sim p^{″})\wedge \mathrm{\forall }{p}^{\prime }\in p,\mathrm{\forall }{p}^{″}\in S\setminus p(\mathrm{\neg }{p}^{\prime }\sim p^{″})\}$

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3: Note

$S/\sim$ is a partition of $S$, which means that each element of $S$ is contained in a single element of $S/\sim$: for each $p\in S$, there is the element of $S/\sim$, $[p]$, defined by $\{p^{\prime }\in S|p^{\prime }\sim p\}$, because for each $p^{\prime },p^{″}\in [p]$, $p^{\prime }\sim p$ and $p^{″}\sim p$, which implies that $p^{\prime }\sim p^{″}$, and for each $p^{\prime }\in [p],p^{″}\in S\setminus [p]$, $p\sim p^{\prime }$ and $\mathrm{\neg }p\sim p^{″}$, which implies that $\mathrm{\neg }{p}^{\prime }\sim p^{″}$, because if $p^{\prime }\sim p^{″}$, $p\sim p^{″}$, a contradiction; $p$ does not belong to any other element of $S/\sim$, because if $p$ belonged to another element, $[p^{\prime }]\in S/\sim$, for each element, $p^{″}\in [p^{\prime }]$, $p^{″}\sim p$, which would mean that $p^{″}\in [p]$, and for each element, $p^{″}\in [p]$, $p^{″}\sim p$, which would mean that $p^{″}\in [p^{\prime }]$, so, $[p^{\prime }]=[p]$ after all.

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References

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