893: For Map, Image of Subset Minus Subset Is Not Necessarily Image of 1st Subset Minus Image of 2nd Subset

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description/proof of that for map, image of subset minus subset is not necessarily 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 is not necessarily 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:
It is not necessarily that $f(S_{1,1}\setminus S_{1,2})=f(S_{1,1})\setminus f(S_{1,2})$
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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, $S_{1,1},S_{1,2}\subseteq S_1$, it is not necessarily that $f(S_{1,1}\setminus S_{1,2})=f(S_{1,1})\setminus f(S_{1,2})$.

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3: Proof

A counterexample suffices. Let $S_1=\{0,1\}$, $S_2=\{0,1\}$, $f:0\mapsto 0,1\mapsto 0$, $S_{1,1}=\{0,1\}$, $S_{1,2}=\{1\}$. Then, $f(S_{1,1}\setminus S_{1,2})=f(\{0\})=\{0\}\ne \mathrm{\varnothing }=\{0\}\setminus \{0\}=f(S_{1,1})\setminus f(S_{1,2})$. Apparently, the non-injectiveness of $f$ prevents the equation from holding.

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4: Note

Compare this with the proposition on preimages.

$f(S_{1,1})\setminus f(S_{1,2})\subseteq f(S_{1,1}\setminus S_{1,2})$, by 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.

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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