Image of the union and intersection of sets. 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.
 A: $I$ is a random set of indices, and the way to prove the statement is the same as for two sets, you'll have to be careful with your arguments if $I$ can be infinite.
Hint for the second part: You'll need injectivity to show that the elements you can find in $S_1$ and $S_2$ are the same.
A: Let $ y\in f(S_{1})\cap f(S_{2}) $. Then $ y\in f(S_{1}) $ and $ y\in f(S_{2}) $. Then there exist $ x_{1}\in S_{1} $ and $ x_{2}\in S_{2} $ such that $ y=f(x_{1})=f(x_{2}) $. Since f is injective we have $ x_{1}=x_{2} $. So, $ y\in f(S_{1}\cap S_{2}) $. Hence $ f(S_{1})\cap f(S_{2}) \subseteq f(S_{1}\cap S_{2}) $.
A: 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 $$ f\left( \bigcup _{i\in I}S_{i}\right)= \bigcup _{i\in I}f(S_{i}). $$
A: For the sake of completeness, the intersection part :
Consider $ f: X \rightarrow Y $. Let $X= A \cup (X-A) $ and define $B=X-A$
then $ A \cap B $ is empty
Assume f is not injective, so that for $ a \in A ; b \in B , f(a)=f(b) $
Then $ f( (A \cap B)= f(\emptyset)= \emptyset \neq f(A) \cap f(B) $
