# 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.

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 $$f\left( \bigcup _{i\in I}S_{i}\right)= \bigcup _{i\in I}f(S_{i}).$$

$I$ 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.

• Hey Henrik. Thank you for the hint. I have to careful in what sense? Feb 3 '15 at 18:32

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 $f(S_{1})\cap f(S_{2}) \subseteq f(S_{1}\cap S_{2})$.

• Can you give me more information on the first problem please? Feb 3 '15 at 19:28
• Ok I'll add another answer for your first question. $I$ is just an index set. An index set is a set whose members label (or index) members of another set. For more details see the following link.en.wikipedia.org/wiki/Index_set
– ASB
Feb 4 '15 at 1:36

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 $$f( (A \cap B)= f(\emptyset)= \emptyset \neq f(A) \cap f(B)$$