191: Map Image of Intersection of Sets Is Contained in Intersection of Map Images of Sets

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A description/proof of that map image of intersection of sets is contained in intersection of map images of sets

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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
• 3: 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 between sets, the map image of any intersection of subsets is contained in the intersection of the map images 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 possibly uncountable number of subsets of $S_1$, $S_{1_{\alpha }}\subseteq S_1$, the map image of the intersection of the subsets, $f({\cap }_{\alpha }{S}_{1_{\alpha }})$, is contained in the intersection of the map images of the subsets, ${\cap }_{\alpha }f(S_{1_{\alpha }})$, which is, $f({\cap }_{\alpha }{S}_{1_{\alpha }})\subseteq {\cap }_{\alpha }f(S_{1_{\alpha }})$.

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2: Proof

For any $p\in f({\cap }_{\alpha }{S}_{1_{\alpha }})$, there is a $p^{\prime }\in {\cap }_{\alpha }{S}_{1_{\alpha }}$ such that $p=f(p^{\prime })$, which means that $p^{\prime }\in S_{1_{\alpha }}$ for every $\alpha$. So, $p\in f(S_{1_{\alpha }})$ for every $\alpha$. So, $p\in {\cap }_{\alpha }f(S_{1_{\alpha }})$.

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3: Note

$f({\cap }_{\alpha }{S}_{1_{\alpha }})={\cap }_{\alpha }f(S_{1_{\alpha }})$ does not necessarily hold as is proved in another proposition.

$f({\cup }_i{S}_{1_i})={\cup }_{i}f(S_{1_i})$ always holds as is proved in another proposition.

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References

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