243: Difference of Map Images of Subsets Is Contained in Map Image of Difference of Subsets

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A description/proof of that difference of map images of subsets is contained in map image of difference of subsets

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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.

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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 contained in the map image of the difference of the subsets.

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

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2: Proof

For any $p\in f(S_3)\setminus f(S_4)$, $p\in f(S_3)$ and $p\notin f(S_4)$, there is a $p^{\prime }\in S_3$ such that $p=f(p^{\prime })$, but $p^{\prime }\notin S_4$, because otherwise, $p=f(p^{\prime })\in f(S_4)$, a contradiction, so, there is a $p^{\prime }\in S_3\setminus S_4$, $f(p^{\prime })=p\in f(S_3\setminus S_4)$.

Why not $f(S_3)\setminus f(S_4)=f(S_3\setminus S_4)$? For any $p\in f(S_3\setminus S_4)$, there is a $p^{\prime }\in S_3\setminus S_4$ such that $p=f(p^{\prime })$, but $f(p^{\prime })$ may be in $f(S_4)$, because as $f$ is not necessarily injective, $f^{-1}(f(S_4))$ is not necessarily $S_4$, so, $p^{\prime }\notin S_4$ could map into $f(S_4)$.

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References

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