Image of intersection of collection of sets - Proof and example

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Let $F :A \rightarrow B$ be a function and $A_i$ with $i \in I$ subsets of $A$.

First show that: $$ F(\cap_{i\in I} A_i) \subset \cap_{i \in I} F(A_i) $$

Then find an example where $\subsetneq$ applies.


Here's my proof:

Let $b \in F(\cap_{i \in I}A_i)$. Therefore $\exists a \in \cap_{i \in I}A_i$ such that $F(a) = b$. Since $a \in \cap_{i \in I}A_i$, it follows that $a \in A_i$ is valid $\forall i \in I$. Since: $$ F(A_i) = \{f(a) | a \in A_i\} $$ we can conclude that $b \in F(A_i) \forall i \in I$, and finally $b \in \cap_{i \in I}F(A_i)$.


I think I was able to prove it but I can't find an example with proper subset... Can someone please give me a hint or show me an example? Also, is my proof correct? If so, what would you change to improve it a.k.a make it more "elegant"?

Thank you!

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The proof looks great, I wouldn't change it.

For counter example, take $f:\{0,1,2\}→\{0,1\}$ with $f(0)=0,f(1)=1,f(2)=1$.

Then $A_0=\{0,1\},A_1=\{0,2\},A_0\cap A_1=\{0\},f(\{0\})=\{0\},f(A_0)=\{0,1\},f(A_1)=\{0,1\},$$f(A_0)\cap f(A_1)=\{0,1\}$ and $\{0\}\subsetneq\{0,1\}$