Let $f:X\to Y$ be a function, and let $\{S_{i}:i\in I\}$ be a family of subsets of $X$. Then, $$f\left(\bigcup_{i
\in I}S_i\right) = \bigcup_{i
\in I}f(S_i).$$
The case where $f(A\cup B)= f(A)\cup f(B)$ is trivial and I've proved this many times in other classes. However, I believe that the problem I am running into is with notation. That is, I don't understand what the set $I$ is. Will my proof method be just the same?
Also, I would like a little help on one more problem. If $S_{1}$ and $S_{2}$ are subsets of a set $X$, and if $f:X\to Y$ is an injection, then $f(S_{1}\cap S_{2})=f(S_{1})\cap f(S_2)$. Now, I know how to prove $f(S_{1}\cap S_{2})\subseteq f(S_{1})\cap f(S_2)$ if our function is not injective, and I know counterexamples of why it isn't equal our function isn't injective. Unfortunately, I am not sure how to use the fact that our function is injective to prove $f(S_{1})\cap f(S_2) \subseteq f(S_{1}\cap S_{2})$. Any help would be much appreciated. Thank you very much!
Note: These questions are coming from Rotman's Intro to Abstract Algebra Chapter 2.
Best Answer
Answer for the first question:
Let $$ y\in f\left( \bigcup _{i\in I}S_{i}\right) $$ $$ \Rightarrow \text{ there exists }x\in \bigcup _{i\in I}S_{i}\text{ such that }f(x)=y $$ $$ \Rightarrow \text{ there exists }x\in S_{i}\text{ for some }i\in I\text{such that }f(x)=y $$ $$ \Rightarrow y\in f(S_{i})\text{ for some }i\in I $$ $$ \Rightarrow y\in \bigcup _{i\in I}f(S_{i}) $$ Therefore $$ f\left( \bigcup _{i\in I}S_{i}\right)\subseteq \bigcup _{i\in I}f(S_{i}). $$ Now let $$ y\in \bigcup _{i\in I}f(S_{i}) $$ $$ \Rightarrow y\in f(S_{i})\text{ for some }i\in I $$ $$ \Rightarrow \text{ there exists }x\in S_{i}\text{ such that }f(x)=y\text{ for some }i\in I $$ $$ \Rightarrow \text{ there exists }x\in \bigcup _{i\in I}S_{i}\text{ such that }f(x)=y $$ $$ \Rightarrow y\in f\left( \bigcup _{i\in I}S_{i}\right) $$ Therefore $$ \bigcup _{i\in I}f(S_{i})\subseteq f\left( \bigcup _{i\in I}S_{i}\right). $$
Hence $$ f\left( \bigcup _{i\in I}S_{i}\right)= \bigcup _{i\in I}f(S_{i}). $$