244: Difference of Map Images of Subsets Is Map Image of Difference of Subsets if Map Is Injective

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A description/proof of that difference of map images of subsets is map image of difference of subsets if map is injective

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of set.
• The reader knows a definition of map.
• The reader admits the proposition that for any map between any sets, the difference of the map images of any subsets is contained in the map image of the difference of the subsets.

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Target Context

• The reader will have a description and a proof of the proposition that for any map between any sets, the difference of the map images of any subsets is the map image of the difference of the subsets if the map is injective.

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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: Description

For any sets, $S_1$ and $S_2$, any injective map, $f:S_1\to S_2$, and any subsets, $S_3,S_4\subseteq S_1$, $f(S_3)\setminus f(S_4)=f(S_3\setminus S_4)$.

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2: Proof

By the proposition that for any map between any sets, the difference of the map images of any subsets is contained in the map image of the difference of the subsets, $f(S_3)\setminus f(S_4)\subseteq f(S_3\setminus S_4)$.

For any $p\in f(S_3\setminus S_4)$, there is a $p^{\prime }\in S_3\setminus S_4$ such that $p=f(p^{\prime })$, so, $p^{\prime }\in S_3$ and $p^{\prime }\notin S_4$. $f(p^{\prime })\in f(S_3)$. $f(p^{\prime })\notin f(S_4)$, because for any $p^{″}\in f(S_4)$, there is a $p^{‴}\in S_4$ such that $p^{″}=f(p^{‴})$, but as $p^{‴}\ne p^{\prime }$, $p^{″}=f(p^{‴})\ne f(p^{\prime })=p$ as $f$ is injective, so, $p$ cannot be a point of $f(S_4)$. So, $p\in f(S_3)\setminus f(S_4)$.

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References

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