292: Map Image of Union of Sets Is Union of Map Images of Sets

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A description/proof of that map image of union of sets is union of map images 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 image of any union of sets is the union of the map images 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 $S_1$, $S_{1_i}\subseteq S_1$, the map image of the union of the subsets, $f({\cup }_i{S}_{1_i})$, is the union of the map images of the subsets, ${\cup }_{i}f(S_{1_i})$, which is $f({\cup }_i{S}_{1_i})={\cup }_{i}f(S_{1_i})$.

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2: Proof

For any element, $p\in f({\cup }_i{S}_{1_i})$, $p\in f(S_{1_i})$ for an i, because as there is an element in ${\cup }_i{S}_{1_i}$ that maps to p, the element has to be in an $S_{1_i}$, which means that $p\in {\cup }_{i}f(S_{1_i})$. For any element, $p\in {\cup }_{i}f(S_{1_i})$, $p\in f(S_{1_i})$ for an i, which means that $p\in f({\cup }_i{S}_{1_i})$, because as there is an element in $S_{1_i}$ that maps to p, the element is also in ${\cup }_i{S}_{1_i}$.

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3: Note

It is important to be aware of that there is no such thing as a "limit element" in ${\cup }_i{S}_{1_i}$, which does not belong to any $S_{1_i}$ but to which a sequence of elements infinitely nears.

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References

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