293: Map Preimage of Union of Sets Is Union of Map Preimages of Sets

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A description/proof of that map preimage of union of sets is union of map preimages of sets

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Topics

About: set
About: map

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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, the map preimage of any union of sets is the union of the map preimages of the sets.

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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 $S_2$, $S_{2_i}\subseteq S_2$, the map preimage of the union of the subsets, $f^{-1}({\cup }_i{S}_{2_i})$, is the union of the map preimages of the subsets, ${\cup }_i{f}^{-1}(S_{2_i})$, which is $f^{-1}({\cup }_i{S}_{2_i})={\cup }_i{f}^{-1}(S_{2_i})$.

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2: Proof

For any element, $p\in f^{-1}({\cup }_i{S}_{2_i})$, $f(p)\in {\cup }_i{S}_{2_i})$, so, $f(p)\in S_{2_i}$ for an i, so, $p\in f^{-1}(S_{2_i})$, so, $p\in {\cup }_i{f}^{-1}(S_{2_i})$. For any element, $p\in {\cup }_i{f}^{-1}(S_{2_i})$, $p\in f^{-1}(S_{2_i})$ for an i, so, $f(p)\in S_{2_i}$, so, $f(p)\in {\cup }_i{S}_{2_i}$, so, $p\in f^{-1}({\cup }_i{S}_{2_i})$.

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3: Note

It is important to be aware of that there is no such thing as a "limit element" in ${\cup }_i{S}_{2_i}$, which does not belong to any $S_{2_i}$ but to which a sequence of elements infinitely nears.

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References

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