894: For Map, Image of Subset Minus Subset Contains Image of 1st Subset Minus Image of 2nd Subset

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description/proof of that for map, image of subset minus subset contains image of 1st subset minus image of 2nd subset

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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 map.

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Target Context

• The reader will have a description and a proof of the proposition that for any map, the image of any subset minus any subset contains the image of the 1st subset minus the image of the 2nd subset.

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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:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$f$: $S_1\to S_2$
$S_{1,1}$: $\subseteq S_1$
$S_{1,2}$: $\subseteq S_1$
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Statements:
$f(S_{1,1})\setminus f(S_{1,2})\subseteq f(S_{1,1}\setminus S_{1,2})$
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2: Natural Language Description

For any sets, $S_1,S_2$, any map, $f:S_1\to S_2$, and any subsets, $S_{1,1},S_{1,2}\subseteq S_1$, $f(S_{1,1})\setminus f(S_{1,2})\subseteq f(S_{1,1}\setminus S_{1,2})$.

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3: Proof

Whole Strategy: Step 1: for each $p\in f(S_{1,1})\setminus f(S_{1,2})$, take a $p^{\prime }\in S_{1,1}\setminus S_{1,2}$ such that $p=f(p^{\prime })$, and conclude the proposition.

Step 1:

For each $p\in f(S_{1,1})\setminus f(S_{1,2})$, there is a $p^{\prime }\in S_{1,1}\setminus S_{1,2}$ such that $p=f(p^{\prime })$, because while $p$ is realized as $p=f(p^{\prime })$ for a $p^{\prime }\in S_{1,1}$, if $p^{\prime }\in S_{1,2}$, $f(p^{\prime })\in f(S_{1,2})$, which could not be $p$ because $p\notin f(S_{1,2})$, so, $p$ is realized by a $p^{\prime }\in S_{1,1}\setminus S_{1,2}$.

So, $p=f(p^{\prime })\in f(S_{1,1}\setminus S_{1,2})$.

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4: Note

The equality holds when $f$ is injective, by the proposition that for any injective map, the image of any subset minus any subset is the image of the 1st subset minus the image of the 2nd subset.

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References

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